EXERCISE.1 ADDING AND RESOLVING VECTORS Q1 (i) 5 + 9 = 14 N north [Vectors are in the same direction, therefore add them.] (ii) 9-5= 4 N north [Vectors are in the opposite direction, therefore subtract them and use the direction of the bigger one.] 1 N (iii) θ R = 1 + 9 Þ R= 144 + 81 = 5 = 15 [Pythagoras] 9 N R 9-1 9 tanq= Þ q= tan = 36.87 [SOHCAHTOA] 1 1 15 N @ 36.87 south of west (iv) R θ 7 N 4 N R = 7 + 4 Þ R= 49 + 576 = 65 = 5 [Pythagoras] 4-1 4 tanq= Þ q= tan = 73.74 [SOHCAHTOA] 7 7 5 N @ 73.74 north of east Adjacent Hypotenuse Opposite sinq= Þ Hypotenuse Opposite = 0( sin30 = ) [SOHCAHTOA] 17.3 N horizontal and N vertical Q (i) cosq= Þ Adjacent = 0( cos 30 = ) 17.3 [SOHCAHTOA] 0 N 0(sin30 ) N 30 0(cos30 ) N Adjacent Hypotenuse Opposite sinq= Þ Hypotenuse Opposite = 5( sin60 = ) 4.33 [SOHCAHTOA].5 N horizontal and 4.33 N vertical (ii) cosq= Þ Adjacent = 5( cos60 = ).5 [SOHCAHTOA] 5 N 5(sin60 ) N 60 5(cos60 ) N Adjacent Hypotenuse Opposite sinq= Þ Opposite = 8 ( sin45 = ) 8 [SOHCAHTOA] Hypotenuse 8 N horizontal and 8 N vertical (iii) cosq= Þ Adjacent = 8 ( cos 45 = ) 8 [SOHCAHTOA] 8 N 45 8 (cos45 ) N 8 (sin45 ) N Adjacent Hypotenuse Opposite sinq= Þ Hypotenuse Opposite = 0( sin30 = ) [SOHCAHTOA] 5.64 N horizontal and.05 N vertical (iv) cosq= Þ Adjacent = 0( cos30 = ) 17.3 [SOHCAHTOA] 6 N 6(sin0 ) N 0 6(cos0 ) N Exercise.1 Page 1
EXERCISE. DISPLACEMENT, VELOCITY AND ACCELERATION s = Þ = Þ = 7 0 = 140 t Q1 v s vt s ( )( ) s = 140 m s t Q v = Þ s = vt Þ s = ( 1)( 10) = 1400 [Change time to SI units and don t forget the Q3 Q4 Q5 Q6 Q7 s = 1 440 m north direction] - 1 40 000 0-1 40 km h = 40 000 m in 3600 s Þ v = = m s [Change velocity to SI units] 3600 9 ( 1500)( 9) s s 1500 v = t t 135 t Þ = v Þ = 0 = 0 = 9 t = 135 s u+ v 8 + 30 average velovity = = = 19 [Don t forget to include the direction if it s given in the question.] average velocity = 19 m s 1 north v- u 15-0 a = = = 5 t 3 a = 5 m s v- u - 30 a = = =-.5 t 8 deceleration =.5 m s [The answer could be given as acceleration =.5 m s 1.] - 1 36 000-1 36 km h = 36 000 m in 3600 s Þ v = = m s [Change velocity to SI units] 3600-1 90 000-1 90 km h = 90 000 m in 3600 s Þ v = = 5 m s [Change velocity to SI units] 3600 v- u 5- a = = = 3 t 5 a = 3 m s s 8 15 = Þ = Þ = 3 365.5 4 60 60 = 9.467 m t s 8 15 v = Þ s = vt Þ s = 3 8.31 60 = 1.4958 m t s 8 16 v = Þ s = vt Þ s = 3 4.37 365.5 4 60 60 = 4.137 m t s 8 6 v = Þ s = vt Þ s = 3.5 365.5 4 60 60 =.37 m t Q8 (i) v s vt s ( )( ) (ii) ( )( ) (iii) ( )( ) (iv) ( )( ) Exercise. Page
EXERCISE.3 UVAST PROBLEMS Q1 v- u 0-60 a = = =- 0.5 t 10 [Change time to SI units and remember aircraft stops v = 0] a = 0.5 m s 1 1 1 1 s = ut + at Þ 1 = 0 t + 8 t Þ t = Þ t = = 0.5 4 a = 0.5 m s Q ( ) ( ) 64-900 v = u + as Þ 8 = 30 + - 5 s Þ s = = 83.6 m - s = 83.6 m v = u + at Þ 8 = 30-8- 30 5t Þ t = = 4.4-5 t = 4.4 s Q3 ( ) ( ) ( ) Q4 ( ) ( )( ) v = u + as Þ 15 = u + 3 36 Þ u = 5-16 = 9 = 3 u = 3 m s 1 v = u + at Þ 40 = + a 8 Þ 40- a = = 3.75 8 a = 3.75 m s 1 1 s = ut + at Þ s = 8 + 3.75 8 = 00 or Q5 (i) ( ) (ii) ( )( ) ( )( ) ( ) - ( ) ( ) 40 1500 v = u + as Þ s = = = 00 3.75 7.5 s = 00 m Exercise.6 Page 3
EXERCISE.4 DISTANCE TIME AND VELOCITY TIME GRAPHS Q1 s / m 15 y- y1 15-0 v= slope Þ v= = = 3 x- x1 5-0 v = 3 m s 1 1 1 Q s = ut + at Þ s ( )( ) ( )( ) 1 = 0 1 + 1 = 1 1 1 s = ut + at Þ s ( )( ) ( )( ) 1 = 0 + = 4 1 1 s = ut + at Þ s ( )( ) ( )( ) 1 = 0 3 + 3 = 9 s / m 9 t / s 5 11 1 4 Q3 (i) 1 v / m s 1 1 3 t / s 0 (ii) v / m s 1 15 19 t / s 0 1 6 15 19 t / s v = 1 m s 1 y- y1 0-0 (iii) a= slope Þ a= = = x- x1-0 a = m s distance = area Þ s = 0 5 = 0 (iv) ( )( ) s = 0 m Exercise.4 Page 4
EXERCISE 3.1 MOMENTUM Q1 p mv ( )( ) = = 65 16 = 40 p = 1 040 kg m s 1 Q p mv mu ( )( ) ( )( ) D = - = 0.43 14-0.43-4 = 7.74 p = 7.74 kg m s 1 Q3 (i) p1= mu 1 1 ( 1)( 3.5) = 4 v = 4 kg m s 1 (ii) mu 1 1+ mu = ( m1+ m) v ( 1)( 3.5) + ( 1)( 0) = ( 1+ 1) v v = 1.75 m s 1 Q4 mu 1 1+ mu = mv 1 1+ mv ( 40)( 0) + ( 50)( 0) = ( 40)( ) + ( 50)( - v ) [Both the child and the boat were initially at rest, which implies u 1 = 0 and u =0. As the child and the boat move in opposite directions as the child steps off, I have chosen to describe the velocity of the boat as v so as to yield a positive value for velocity.] 80 v = = 1.6 50 v = 1.6 m s 1 [You may wish to include the direction, in which the answer would be v = 1.6 m s 1 away from the pier or in the opposite direction to the boy.] Q5 mu 1 1+ mu = mv 1 1+ mv ( 0.019)( 0) + ( 1.35)( 0) = ( 0.019)( 460) + ( 1.35)( - v ) [Remember to put the mass of the bullet into SI units. As the bullet and the gun move in opposite directions when the bullet is fired, I have chosen to describe the velocity of the gun as v so as to yield a positive value for velocity.] ( 0.019)( 460) v = = 6.47 1.35 v = 6.47 m s 1 mu + mu = m + m v Q6 ( ) 1 1 1 ( 50)( 3) + ( 75)( 0) = ( 50 + 75) v 150 v = = 1. 15 v = 1. m s 1 Exercise 3.1 Page 5
Q7 mu 1 1+ mu = mv 1 1+ mv Q8 ( 4)( 4) + ( )( - 3) = ( 4)( 1) + ( ) v [The spheres are initially moving in opposite directions, therefore must have opposite signs, i.e. sphere 1 moves in the positive direction and sphere moves in ( ) + (- )- ( ) 16 6 4 6 v = = = 3 v = 3 m s 1 the negative direction.] This question is set so as challenge the best students, but may be beyond the level required at Leaving Cert. ( ) mu 1 1+ mu = m1+ m v1 [Deal with momentum in the east west direction first.] ( )( 6.1) + ( 0.035)( 0) = (.035) v 1 [Remember to put the mass of the bullet into SI units.] 1.440 v 1 = =.035 407 ( ) mu 1 1+ mu = m1+ m v [Deal with momentum in the north south direction next.] ( )( 0) + ( 0.035)( 350) = (.035) v 1.5 450 v = =.035 407 æ440ö æ450ö v= ( v1) + ( v) = + = 7.177 = 8.5 ç è407 ø çè407 ø v = 8.5 m s 1 Exercise 3.1 Page 6
EXERCISE 3. WEIGHT AND GRAVITY Q1 Q (i) (ii) - 11 4 ( )( )( ) 8 ( 3.8 ) 37 Gm1m 6.7 6 7.814 F = = = = 1.95 17 d 1.444 F = 1.95 0 N ( - 11)( ) 6 ( 1.7 ) 1 GM 6.7 7 4.69 g = = = = 1.6 1 R.89 g = 1.6 m s ( - 11)( 3) 6 ( 3.4 ) 13 GM 6.7 6.4 4.88 g = = = = 3.7 13 R 1.156 g = 3.7 m s Q3 ( ) ( ) ( ) 65 v = u + as Þ 0 = 5 + - 1.6 s Þ s = = 195.3 3. h = 195.3 m 65 v = u + as Þ ( 0) = ( 5) + ( - 9.8) s Þ s = = 31.9 19.6 h = 31.9 m 1 1 1.6 s = ut + at Þ 1.6 = 0 0.6 + g 0.6 Þ g = = 8.9 0.18 g = 8.9 m s Q4 ( )( ) ( ) Q5 F F midwife Mars - 11 ( 6.7 )( 7)( 4) ( 0.45) Gm1m = = = d - 11 3 ( 6.7 )( 6.4 )( 4) ( 5.6 ) - 8 9.53 N Gm1m = = = d [When the stone is at its greatest height its velocity is zero.] - 8 5.47 N F midwife > F Mars The midwife exerts a stronger gravitational attraction. 0 Exercise 3. Page 7
Q6 (i) W mg W ( )( ) Q7 = Þ = 100 9.8 = 11760 W = 11 760 N (ii) W1 11760 W = = = 1306.67 9 9 [Force due to gravity is inversely proportional to the square of the distance between the centres of gravity of the bodies in question. Therefore, if the distance is tripled, then the force is one ninth of its original value.] W = 1 306.67 N (iii) m = 1 00 kg W W - 11 4 ( 6.7 )( 6 )( 80) 6 ( 6.371 ) - 11 4 ( 6.7 )( 6 )( 80) 6 ( 6.38 ) GMm = = = 79.3 d sea level GMm = = = 789.59 d altitude D W= Wsea level - Waltitude = 79.3-789.59 =.73 W =.7 N [The mass of a body does not change regardless of location.] Exercise 3. Page 8
EXERCISE 3.3 FORCE Q1 (i) p mv mu ( )( ) ( )( ) Q (ii) D = - = 400 80-400 60 = 48 000 p = 4.8 4 kg m s 1 v- u 80-60 a = = =.5 t 8 a =.5 m s (iii) F ma ( )( ) = = 400.5 = 6 000 F = 6 000 N 5 5 ( )( )- ( )( ) 1.6 150 1.6 50 7 mv - mu 1.6 F = = = = 4 t 40 40 F = 4 5 N Q3 F = ma Þ T - Fr = ma Þ T = ma + Fr ( )( ) ( ) T = 4.5 0.6 + 8 = 3.5 F = 3.5 5 N 5 4 5-36 Q4 ( ) ( ) ( )( ) v = u + as Þ 0 = 6 + a 7. Þ a = =-.5 14.4 ( )( ) F = ma Þ F r = 0.6.5 = 1.5 [The negative sign associated with the acceleration can be ignored as we are only interested in the magnitude of the force.] F = 1.5 N - 64 Q5 ( ) ( ) ( )( ) v = u + as Þ 0 = 8 + a 0.04 Þ a = =- 800 0.08 ( )( ) F = ma Þ F r = 0 800 = 80 000 [The negative sign associated with the acceleration can be ignored as we are only interested in the magnitude of the force.] F = 8 4 N 5 Exercise 3.3 Page 9
EXERCISE 3.4 RESOLVING FORCES Q1 R θ 48 N 14 N R = 48 + 14 Þ R= 304 + 196 = 50 [Pythagoras] 14-1 14 tanq= Þ q= tan = 16.6 [SOHCAHTOA] 48 48 50 N @ 16.6 north of east Q 1 kn 1(sin30 ) kn 30 1(cos30 ) kn Adjacent cosq= Þ Hypotenuse Adjacent = 1( cos30 = ) 0.866 [SOHCAHTOA] Opposite sinq= Þ Hypotenuse Opposite = 1( sin30 = ) 0.5 [SOHCAHTOA] 0.866 kn horizontal & 0.5 kn vertical Q3 30 N R θ 80 N R = 80 + 30 Þ R= 6 400+ 900 = 85.44 [Pythagoras] 30-1 30 tanq= Þ q= tan = 0.56 [SOHCAHTOA] 80 80 R = 85.44 N @ 0.56 north of west [When a third force is added, in order for the resultant of the three forces to be equal to zero, the third force must be equal in magnitude and opposite in direction to the resultant of the first two forces.] F = 85.44 N @ 0.56 south of east Q4 (i) R F r ( 50g ) cos0 0 ( 50g ) sin0 ( )( ) T = mgsinq= 50 9.8 sin0 = 167.6 [Resolving the weight: labelled here as T but there is no significance to that, any notation could be used.] F = ma Þ T - Fr = ( 50)( 0) Þ Fr = T [In order for the acceleration to be zero, the friction force must be equal to the component of weight parallel to the plane.] F r = 167.6 N (ii) F = ma Þ T - Fr = ( 50) a Þ 167.6- ( 0) = 50a Þ a = 3.35 a = 3.35 m s 1 1 (iii) s = ut + at = ( 0)( ) + ( 3.35)( ) = 6.7 s = 6.7 m Exercise 3.4 Page
EXERCISE 4.1 MOMENTS OF FORCES Q1 M Fd ( )( ) = = 40 0. = 8 M = 8 N m 30 M = Fd Þ 30 = F 0.15 Þ F = = 00 0.15 F = 00 N Q ( ) Q3 T Fd T ( )( ) = Þ = 0 0.07 = 1.4 [Remember that with couples, the value T = 1.4 N m T = Fd Þ 6 = F 0.5 Þ F = 1 F = 1 N Q4 ( ) ( ) Q5 d is the distance between the lines of action of the forces.] 0 35 6 N 6 N x 6 N Anticlockwise moments = clockwise moments ( 6)( 30) + ( 6)( 15) = ( 6)( x) [Moments are found about the 50 cm x = 45 Weight is at the 95 cm mark. mark because the force due to the tension in the rope is not known, therefore the moment due to that force must be made zero by taking the moments about a point along its line of action.] Exercise 4.1 Page 11
EXERCISE 4. DENSITY Q1 V = ( 1.5) 3 = 3.375 m m r = Þ 85= Þ m = ( 85)( 3.375) = 86.875 V 3.375 m = 86.875 kg Q V = ( 50)( 0)( 1.8) = 1800 m m r = Þ 1 = Þ m = ( 1 )( 1800) = 1.8 V 1800 m = 1.8 6 kg 3 3 6 m 5-3 Q3 r = = = 0.8 g cm V 65 6-3 0.8-3 0.8 1-3 - 3 0.8 g cm Û kg cm Û kg cm Û 0.8 00 kg cm 00 00 [There are 1 000 g in 1 kg and 1 000 000 cm 3 in 1 m 3.] ρ = 800 kg m 3 Exercise 4. Page 1
EXERCISE 4.3 PRESSURE F 550 6 Q1 p = = = 5.5-4 A 1 p = 5.5 6 Pa Q Q3 3 F 17.6 p= = =. - 4 A 80 p =. 6 Pa 6 ( 9.8) F mg 5 m p = 1.013 14 A = A Þ = 5.1 ( 1.013 5 )( 5.1 14 ) Þ m = = 5.7 9.8 m = 5.7 18 kg 18 3 4 p = gh Þ 4 = 00 9.8 h Þ h = = 0.7803 3 Q4 r ( )( )( ) 3 ( 13.6 )( 9.8) h = 780.3 mm Q5 p= pwater + patm [Pressure at bottom of lake due to water plus atmospheric pressure] pwater ( )( )( ) = r gh = 00 9.8 5 = 4.9 4 5 5 p = 4.9 + 1.013 = 1.503 p = 1.503 5 Pa 4 Q6 p p p rgh rgh rg( h h ) difference = 1 - = 1 - = 1 - [Difference in pressure due to different depths] p difference = ( 00)( 9.8)( 0.3) = 940 P difference = 940 Pa Q7 p r gh ( 00)( 9.8)( 94) ( 1.013 5 ) = = + [The atmospheric pressure has no significant influence at this depth.] 8 5 1.07055 1.013 1.071565 p = + = p = 1.07 8 Pa Exercise 4.3 Page 13
EXERCISE 4.4 BOYLE S LAW Q1 pv 1 1 pv = [Boyle s law] ( 13) V = ( 53.5) 4V = V 1 V 1 Volume is increased by a factor of 4. Q pv 1 1= pv ( 5 )( ) = ( 5 ) 1.01.7 1.1 V ( )( ) 5 1.01.7 5.77 V = = =.5 5 5 1.1 1.1 3 V =.5 cm Q3 (i) pv 1 1= pv ( 5 ) pv = 1.01 V [ doubles in size V = V 1 ] 1 1 1 5 p 1 =.0 Pa (ii) p1 pwater patm = + [Pressure at bottom of lake due to water.0 = + 1.01 5 5 p water plus atmospheric pressure] 5 p water = 1.01 Pa (iii) p = r gh [Pressure in a fluid] ( )( ) 5 1.01 = 00 9.8 h 5 1.01 505 h = = =.306 9800 49 h =.3 m Exercise 4.4 Page 14
EXERCISE 5.1 ENERGY CONVERSIONS Q1 EP = mgh [Potential energy] ( )( )( ) E = 80 9.8 6.14 + 0.3 = 5048.96 P E p = 5 048.96 J Q 1 Ek = mv [Kinetic energy] 1 ( 360 000 )( 80 ) 1.15 9 E k = = E k = 1.15 9 J 1 1 + = + [Principle of Conservation of Energy] 1 ( )( ) ( ) 1 m 9.8.5 + m 0 = m( 9.8)( 0) + mv [height off the ground = 4 1.5; initial velocity = 0; max velocity is when gymnast is at her lowest point.] Q3 mgh1 mu mgh mv v 4.5= Þ v = 49 = 7 v = 7 m s 1 1 1 Q4 (i) mgh1 + mu = mgh + mv [Principle of Conservation of Energy] - 1 ( 3-7.5 )( 9.8)( 370) + m( 0) = ( 7.5 3 )( 9.8)( 0) + Ek Þ Ek = 7.195 E k = 7.195 J 1 (ii) Ek = mv [Kinetic energy] 1 ( ) - 7.195= 7.5 v Þ v= = - 3 7.5 75 = 14 37 = 85.159 v = 85.16 m s 1 Exercise 5.1 Page 15
1 1 Q5 (i) mgh1 + mu = mgh + mv [Principle of Conservation of Energy] 1 1 ( 0.4)( 9.8)( 1.6) + m( 0) = ( 0.4)( 9.8)( 0) + ( 0.4) v Þ v= 31.36 = 5.6 v = 5.6 m s 1 E = mgh = 0.4 9.8 1.6 = 6.7 [Ball at initial height of 1.6 m] (ii) ( )( )( ) pbefore before ( )( )( ) Ep bounce = mghbounce = 0.4 9.8 1. = 4.704 [Ball reaches a height of 1. m on first bounce] E = E - E = 6.7-4.704 = 1.568 p p p lost lbefore lbounce E p lost = 1.568 J (iii) Turned into heat or sound ( )( ) 1 4.704 (iv) mv = 4.704 Þ v = 0.4 = 3.51 = 4.848 v = 4.85 m s 1 [There is always some energy converted to heat during an energy conversion and generally also some lost as sound.] Exercise 5.1 Page 16
EXERCISE 5. POWER Q1 (i) None (ii) W mgh ( )( )( ) [There is the same increase in the boy s potential energy on each occasion, therefore the same amount of work is done on each occasion.] = = 40 9.8 6 = 35 [The work done (by the boy) is equal to the gain in his potential energy.] W 35 P1 = = = 117.6 [Power when walking] t 0 1 W 35 P = = = 94 [Power when running] t 8 P 1 = 117.6 W, P = 94 W Q (i) work = ( total weight)( height) [weight = mg, work = mgh] ( ( ))( ) W = 7900 + 3 700 45 = 450 000 W = 4.5 5 J (ii) 5 W 4.5 P = = = t 60 P =7.5 kw 7500 Po 0 7.5 0 1 Percentage efficiency = = = 83 Pi 9 3 Percentage efficiency = 83.33% Exercise 5. Page 17
EXERCISE 6.1 CIRCULAR MOTION 0 Q1 v= wrþ 0 = w( ) Þ w= = [Definition of angular velocity] Q (i) ω = rad s 1 5 T = p p p p 1. w w Þ = w Þ = 1. = 3 [You should note that it may lead to an error if you sub π = 3.14 into an equation, occasionally that approximation can cause a rounding error. It s better to leave the answer in terms of π or use the π button on the calculator if a decimal answer is required.] w= 5p rad s 3-1 p (ii) v wr æ ç ö ( 1.5) 5 5p = = ç = çè3 ø (iii) v = a 5 m s p - 1 v r 5p ( ) = = = 41.13 [It would have been just as acceptable to 1.5 use the centripetal acceleration formula involving angular velocity: 5p a= w r= ( 3 ) ( 1.5) = 41.13 ] a = 41.13 m s - ( )( ) 5p mv 4 (iv) F = = = 164.49 r 1.5 F = 164.5 N = = 1 0. =.4 Q3 v wr ( )( ) v =.4 m s 1 Exercise 6.1 Page 18
Q4 6.8 T = = 0.68 [Periodic time is the total time divided by the number of revolutions (or oscillations).] p( ) p( ) pr 0.3 0.3 T = Þ 0.68 = Þ v= v v 0.68 = 3 [We first need to calculate the velocity.] v = 3.0 m s 1 mv ( 0.05)( 3) = =1.5=F r 0.3 [The tension in the string is 1.5 N; this provided the centripetal force. By calculation we have shown this to also be 1.5 N. Thus we have shown that the experiment verified the formula.] Q5 (i) 1 1 mgh1 + mu = mgh + mv [Principle of Conservation of Energy] 1 ( )( )( ) ( )( ) 1 0.05 9.8 00 + 0.05 6 = ( 0.05)( 9.8)( 0.6) + ( 0.05) v 0.9 = 0.94 + ( 0.05) v Þ v= 0.9-0.94 = 0.05 4.4 = 4.93 v = 4.9 m s 1 (ii) ( ) v 4.93 a = = = 40.4 r 0.6 a = 40.4 m s ( )( ) mv 0.05 4.93 (iii) F = = =.0 r 0.6 F =.0 N Exercise 6.1 Page 19
EXERCISE 6. GRAVITY Q1 Q T r 4π R = R= GM 3 3 T GM 4π ( ) - 4 60 60 ( 6.7 11 )( 6 4 ) [Kepler s third law] 3 + h= [Radius of Orbit = Radius of Earth + 4p 4 6 3 6.4 + h = 3 4p h = 7.6-6.4 = 4.36-6.4 = 3.596 h = 3.6 7 m 3 6 7 6 7 Height above Earth; Periodic time = 1 day 4 hours, 60 minutes per hour, 60 seconds per minute] a= ω r [Definition of centripetal acceleration] 9.8 = ω ( 5) ω = 9.8 = 1.96 = 1.4 5 a = 1.4 rad s 1 Q3 Q4 Q5 3 3 4π R 4π R T = T = [Kepler s third law] GM GM 3 π ( ) 11 4 ( 6.7 )( 6 ) 8 4 3.84 7.35 T = = = 5.56 =.358 14 4.0 T =.36 6 s (7.3 days) T 4π R = R= GM 3 3 T GM 4π ( ) ( )( ) 1 6 [Kepler s third law] 6 11 30 374 6.7.0 37 1.874 3 R = = 3 = 4.748 = 7.80 4π 4π R = 7.8 11 m T 3 35 11 3 3 4π R 4π R = M= [Kepler s third law] GM T G 3 π ( ) 5 11 ( 3.07 ) ( 6.7 ) 8 4 6.7 8 1.187 M = = = 1.880 6.315 M = 1.88 7 kg 7 Exercise 6. Page 0
EXERCISE 7.1 SIMPLE HARMONIC MOTION Q1 F = ks [Hooke s law] ( ) 0.3 =k 0.019 [The reason the force is negative is that it 0.3 300 k = = = 1.789 0.019 19 k = 15.79 N m 1 Alternative approach is in the opposite direction to the displacement; it would be equally correct to have a positive force and negative displacement. The displacement is put into SI units.] = ( l l ) = F l l = 0.3 = 0.3 = 300 F k = 0 k 1.789 0 0.181 0.16 0.019 19 [Rather than worry about the direction (plus or minus) of the force and displacement, it may be preferable to use the formula F = k ( l l 0 ). The displacement still has to be put into SI units.] Q mg =ks [Hooke s law: note that the restoring force is equal in magnitude to the weight but acts in the opposite direction.] ( )( ) 70 9.8 686 ( 70)( 9.8) =k( 0.005) k = = = 13700 0.005 0.005 k = 1.37 5 N m 1 Exercise 7.1 Page 1
Q3 (i) A = cm [Definition of amplitude of oscillation] k k (ii) F =ks ma =ks a = s a = s [The way the question is phrased implies m m 6 60 a = ( 0.1) = = 7.5 0.08 8 a = 7.5 m s that we are interested only in the magnitude of the acceleration.] [The mass and displacement both have to be put into SI units.] k 6 (iii) ω = = = 5 3 m 0.08 π π T = = = 0.76 ω 5 3 T = 0.73 s [Combining the definition of SHM a= ω swith Hooke s law F = ks k we get ω = m ] Q4 (i) (ii) a= ω s [Definition of SHM] 4 4 =ω ( 0.8) ω = ω = 5 0.8 ( ) ( )( ) [The reason the displacement is negative is that it is in the opposite direction to the acceleration; it would be equally correct to have a positive displacement and negative acceleration. The displacement is put into SI units.] a =ω 0. = 5 0. = 1 [Calculating the acceleration when the displacement is 0 cm; the constant of proportionality remains 5 N m 1.] a = 1 m s a= ω s [Definition of SHM] = s s = = 0.4 [Calculating the displacement when the ( ) ω s = 40 cm 5 acceleration is 0 cm; the acceleration is negative because it s in the opposite direction to the displacement.] [Obviously it s perfectly acceptable to leave the answer as 0.4 m.] Q5 T = π l g [Periodic time of a simple pendulum] Exercise 7.1 Page
1. T = π =.199 9.8 T =. s Exercise 7.1 Page 3
EXERCISE 9.1 HEAT CAPACITY AND LATENT HEAT Q1 Q Q3 (i) (ii) Q = mc θ [Specific heat capacity] 80 80 = ( 0.16) c( 0) c = = 56.5 0.16 0 c = 56.5 J kg 1 K ( )( ) Q = mc θ [Specific heat capacity] Q = 3 5 00 0 = 3 5 180 = 75400 ( )( )( ) ( )( )( ) E = 75 400 J θ = 0 0 = 80 θ = 80 C Q = mc θ [Specific heat capacity] Q = 0.5 4180 80 = 16700 ( )( )( ) Q = 167 00 J W (iii) P = W = Pt t W = 000 1 [It s a kw heater; time is 1 s as we are ( )( ) W = 000 J Q 16700 (iv) Q = Wt t = = = 83.6 W 000 t = 83.6 s Q4 ( )( ) asked for the energy per second.] [Total energy is equal to the energy output per second multiplied by the time.] Ep = mgh = m 9.8 600 = 5880m [Potential energy of the water at the top of the waterfall] Ep = Q [The potential energy is converted to heat.] 5880 94 Q = mc θ 5880m = m ( 4180) θ θ = = = 1.407 4180 09 θ = 1.41 C Exercise 9.1 Page 4
Q5 Q6 Q7 Q8 Q= VIt [Formula found by combining Joule s law and Ohm s law, as described on pp. 306 307, also noted with reference to Mandatory Exp 9: Measurement of Specific Heat Capacity, p.138] Q = ( 1)( 6)( 5 60) = 1600 [Time has to be put in SI units; 5 minutes = 5 60 seconds.] Q = mc θ [The energy from the electricity heats the metal.] 1600 1600 = ( 0.8)( 440) θ θ = = 61.363 ( 0.8)( 440) θ = 61.36 C Q= m l [Specific latent heat] Q = ( 0.005)( 3.3 5 ) = 1650 [Mass has to be converted to SI units.] E = 1 650 J Q1 = mc θ [The first step is to cool the water from 17 C to 0 C.] Q 1 = ( 0.1)( 4180)( 17 0) = 76 Q = m l [The next step is to freeze that water at 0 C to ice at 0 C.] 5 4 Q = 0.1 3.3 = 3.3 ( )( ) Q= Q1 + Q [The last step is to add the heat needed to cool the water to the heat needed to change its state.] 4 Q = 3.3 + 76 = 406 Q = 40 6 J Q1 = mc θ [The first step is to heat the water from 0 C to 0 C.] Q 1 = ( 0.)( 4180)( 0 0) = 66 880 Q = m l [The next step is to boil g of that water at 0 C to steam at 0 C.] 6 Q = 0.01.3 = 3000 ( )( ) Q= Q1 + Q [The last step is to add the heat needed to heat the water to the heat needed to change its state.] Q = 66 880 + 3000 = 89 880 Q = 89 880 J Exercise 9.1 Page 5
EXERCISE.1 THE DOPPLER EFFECT Q1 fc f ' = c u [Use the minus version of the Doppler Effect formula because the train is approaching the observer.] ( )( ) ( ) u ( )( ) ( )( ) 640 340 640 340 640 340 70 = 340 u= u= 340 = 340 30. = 37.778 340 70 70 u = 37.78 m s 1 Q fc f ' = c u [Use the minus version of the formula because the car is approaching the Garda.] 500 0050 = 9 8 (.5 )( 3.0 ) 8 ( 3.0 ) u [GHz means gigahertz, giga means 9 ; EM waves are light and therefore travel at 3.0 8 m s 1 ] 9 8 9 8 (.5 )( 3.0 ) (.5 )( 3.0 ) 8 8 3.0 u= u = 3.0 500 0050 500 0050 = 30 u = 30 m s 1 Q3 fc f ' 1 = c u [First consider the observed pitch as the car approaches the observer.] ( )( ) ( ) ( ) 10 340 40800 3 f ' 1 = = = 143 [It is better to leave numbers as decimal 340 55 85 19 during calculations as the decimal 143.158 is 3 only an approximate value for 143 ] 19 f ' fc = [Next consider the observed pitch as the c + u car moves away from the observer.] ( )( ) ( ) + ( ) 10 340 40800 3 f ' = = = 3 340 55 395 79 3 3 f = f' 1 f ' = 143 3 = 39.87 [The apparent change in pitch as the car 19 79 moves past equals the difference between the apparent frequency as it approaches and the apparent frequency as it moves away.] f = 39.9 Hz Exercise.1 Page 6
Q4 (i) c 3.0 c= fλ f = f = λ λ 8 3.0 λ' = 411.54 f ' = = 7.897 9 411.54 8 9 14 8 3.0 λ = 4.17 f = = 7.3140 9 4.17 The star is moving away. 9 14 [We know this because the apparent frequency is greater than the emitted frequency.] (ii) fc f ' = c + u [Use the plus version of the formula because the distant galaxy is moving away from the Milky Way.] 14 7.897 = 14 8 ( 7.3140 )( 3.0 ) 8 ( 3.0 ) u 14 8 ( 7.3140 )( 3.0 ) [GHz means gigahertz, giga means 9 ; EM waves are light and therefore travel at 3.0 8 m s 1.] 3.0 u= u = 3.0 3.00 = 1.0 14 7.897 u = 1 6 m s 1 8 8 8 6 Exercise.1 Page 7
EXERCISE 11.1 VERNIER SCALE ON A VERNIER CALLIPERS Q1 Q Step A: The line on the main scale that is directly in front of the first line of the Vernier scale is 7. Step B: The line on the Vernier scale that lines up best with a line on the main scale is 3. Step C: Combine these two readings to give a reading of 7.3. The reading is 7.3 mm. [You may make the assumption that the Vernier callipers has a unit of measure of the millimetre.] Step A: The line on the main scale that is directly in front of the first line of the Vernier scale is. Step B: The line on the Vernier scale that lines up best with a line on the main scale is 5. Step C: Combine these two readings to give a reading of.5. The reading is.5 mm. [You may make the assumption that the Vernier callipers has a unit of measure of the millimetre.] Questions involving calculating distance using the Vernier scale on a Vernier callipers should not be asked in the Leaving Cert. examination. They are, however, of practical use, as a Vernier callipers may be used while carrying out experiments in the laboratory. Exercise 11.1 Page 8
EXERCISE 11. HARMONICS λ = = =.4 = 4.8 Q1 (i) l λ l ( ) [The pipe is open at both ends means that there is an antinode at both ends. The distance between consecutive antinodes is half the wavelength of the travelling wave.] c 340 45 c= fλ f = = = = 70.833 [The frequency can be worked out because λ 4.8 6 we know the speed of sound.] f = 70.83 Hz (ii) f = f = 141.667 [The second harmonic is two times the fundamental frequency.] f = 114.67 Hz (iii) f3 = 3f = 1.5 [The third harmonic is three times the fundamental frequency.] f 3 = 1.5 Hz λ = = 4 = 4 0.95 = 1.18 4 Q l λ l ( ) c 340 c= fλ f = = = 88.1 λ 1.18 f = 88 Hz [The pipe is closed at one end means that there is an antinode at one end and a node at the other. The distance between a node and adjacent antinode is a quarter of the wavelength of the travelling wave.] f = 3f = 864.4 [The next lowest harmonic is the third harmonic as only odd harmonics are present in pipes closed at one end.] f = 864 Hz Q3 f = 1 T l µ = 1 700 ( ) = 91.56 f = 3.7 1.4 0.0835.8 f = 3.7 Hz [Fundamental frequency of a stretched string] Exercise 11. Page 9
Q4 f = 1 T l µ 1 83 39.77 = 0.66 µ ( ) 83 = ( 1.3)( 39.77) = 435.964 µ 83 = ( 435.964) = 1.895 µ 83 4 µ = = 4.38 5 1.895 µ = 4.38 4 kg m 1 5 Exercise 11. Page 30
EXERCISE 11.3 SOUND INTENSITY Q1 6 W 4 P = = = 4 t 1 6 P 4 I = = = 8 A 0.5 I = 8 6 W m Q ( ) 6 6 [Power is work divided by time, in this case the time is one second.] A= 4πr = 4π 5 = 0π [The area that the sound is passing P 0.0 I = = = 6.366 A 0π I = 6.37 5 W m Q3 A r ( ) 5 = 4π = 4π 0 = 1600π through is equal to the curved surface area of a sphere of radius 5 m; formula on p. of Formula and Tables booklet.] 9 P 13 I = = = 3.97 A 1600π 3.97 13 W m is lower than the threshold of hearing of 1 1 W m, which means the sound cannot be heard by a person with average hearing at that distance. Q4 Increase in I.L. = 3 db [A full explanation of the mathematics of this is not part of the Leaving Cert. syllabus. The explanation as to why halving or doubling the sound intensity causes a decrease or increase of 3 db is given in the derivation of formulae document that is available on the Gill and Macmillan website: http://www.gillmacmillan.ie/additionalresources/] Q5 3 = = 8 [An increase in sound intensity level of 9 db means that the sound intensity has doubled 3 times.] The sound intensity has increased by a factor of 8. Exercise 11.3 Page 31
EXERCISE 1.1 SPHERICAL MIRRORS 1 1 1 1 1 1 Q1 = + = + f u v 15 40 v 1 1 1 1 1 = = v = 4 v 15 40 v 4 v = 4 cm; image is real v 4 3 m = = = = 0.6 u 40 5 m = 0.6 [Concave mirror implies focal length is positive] [We know the image is real since the value of v is positive.] Q 1 1 1 1 1 1 = + = + f u v 5 0 v [Concave mirror implies focal length is positive] 1 1 1 1 4 5 1 = = = v =0 v 5 0 v 0 0 0 v 0 m = = = 5 u 0 [In this case u and v are object and image distances. The value of v is positive as it is a distance (and distances are always positive). v = 0 tells us that the image distance is 0 cm and that the image is virtual.] v v m= 5 = v= 5 3 = 15 u 3 [In this case u and v are object and image heights.] v = 15 cm Q3 1 1 1 1 1 1 = = f u v 0 1 v [Convex mirror implies focal length is positive and that the image is always virtual (i.e. v is negative)] 1 1 = + 1 1 = v = 7.5 v v 15 v = 7.5cm; image is virtual v 7.5 5 m = = = = 0.65 u 1 8 m = 0.65 [The image formed in a convex mirror is always virtual.] [In this case u and v are object and image distances.] Exercise 1.1 Page 3
v v Q4 m= 4= v= 4u u u 1 1 1 1 1 1 = + = + f u v f u 4u 1 4 1 1 5 = + = 4u= 10 u= 30 4 4u 4u 4 4u u = 30 cm 1 1 1 1 1 1 = = f u v f u 4u 1 4 1 1 3 = = 4u= 7 u= 18 4 4u 4u 4 4u u = 18 cm v 1 Q5 m= m= m= u 4 v 1 v 15 m= = v= = 7.5 u 15 (i) Image is real implies mirror is concave 1 = 1 + 1 1 = 1 + 1 1 = 3 f = 15 = 5 f u v f 15 7.5 f 15 3 In the case of the concave mirror, f = 5 cm (ii) Image is real implies mirror is concave [In this case u and v are object and image distances. We don t know what their values are but we can relate them in this way.] [There are two possible cases where the image size will be four times the object size, one where the image is real and the other where it is virtual.] [First look at the case where the image is real.] [Then look at the case where the image is virtual.] [In this case u and v are object and image heights.] [In this case u and v are object and image distances.] [We can make this statement because the image is always virtual in a convex mirror. If we apply the formula before we make this statement, we find that the value of f is positive and it therefore follows that the mirror is concave.] [We cannot assume whether the mirror is concave or convex as both possibilities could lead to virtual diminished images; we therefore just apply the formula and check if the value of f is positive or negative.] Exercise 1.1 Page 33
1 1 1 1 1 1 1 1 = = = f u v f 15 7.5 f 15 In the case of the convex mirror, f = 15 cm Exercise 1.1 Page 34
EXERCISE 13.1 REFRACTIVE INDEX real depth 4.1 Q1 n = = apparent depth 3 apparent depth 6.3 apparent depth = = 1.575 4 apparent depth = 1.575 m [This is the appropriate formula for refractive index.] sini sin39 0.69 Q (i) n = = = = 1.50 sinr sin4.8 0.419 n = 1.5 (ii) The light would be refracted less, r closer to i. 1 3 9 (iii) wng = wna ang = ang = 1.5 = n 4 8 w n g = 1.15 a w [This is the appropriate formula for refractive index.] [This is because the refractive index of water is much closer to the refractive index of glass.] [The third harmonic is three times the fundamental frequency.] Q3 (i) sini sin39.9 0.641 n = = = = 1.53 sinr sin4.8 0.419 [This is the appropriate formula for refractive index.] n = 1.53 (ii) [The refractive index of water is 1.33, which is close to the refractive index of the prism. This means that the light will only be refracted by a small amount.] Q4 8 c1 1 3.0 n= = c = 1.5 c 5 c c = 1.5 8 m s 1 8 [This is the appropriate formula for refractive index.] Exercise 13.1 Page 35
EXERCISE 13. TOTAL INTERNAL REFLECTION 1 1 Q1 n = 1.55 = [This is the appropriate formula for sinc sinc refractive index.] 1 1 1 sinc = C = sin = 40.178 1.55 1.55 C = 40.18 Q1 1 1 1 n = = = = 1.309 sinc sin49.81 0.7639 [This is the appropriate formula for refractive index.] n = 1.31 Q3 n= 1 = sinc = C = 48.59 sinc 3 sinc 4 [This is the appropriate formula for refractive index.] tanc = 1.1339 r r tanc = 1.1339 = r = 17.008 15 15 [This can be seen from Fig 13.14 on p.0 of the textbook.] r = 17 m Exercise 13. Page 36
EXERCISE 13.3 LENSES 1 1 1 1 1 1 Q1 = + = + f u v 0 30 v 1 1 1 1 1 = = v = 60 v 0 30 v 60 v = 60 cm; image is real v 60 m = = = u 30 m = [Converging lens implies focal length is positive.] [We know the image is real since the value of v is positive.] 1 1 1 1 1 1 Q = + = + f u v 15 v 1 1 1 1 1 = = v =30 v 15 v 30 v 30 m = = = 3 u v v m= 3= v= 3 = 6 u v = 6 cm [Converging lens implies focal length is positive] [In this case u and v are object and image distances. The value of v is positive as it is a distance (and distances are always positive). v = 30 tells us that the image distance is 30 cm and that the image is virtual.] [In this case u and v are object and image heights.] 1 1 1 1 1 1 Q3 = = f u v 30 v 1 1 1 1 4 = + = v = 7.5 v 30 v 30 v = 7.5 cm; image is virtual v 7.5 3 m = = = = 0.75 u 4 m = 0.75 [Diverging lens implies focal length is negative] [The image formed in a convex mirror is always virtual.] [In this case u and v are object and image distances.] Exercise 13.3 Page 37
v v Q4 m= = v= u u u [In this case u and v are object and image distances, we don t know what their values are but we can relate them in this way.] [There are two possible cases where the image size will be twice the object size: one where the image is real and the other where it is virtual.] 1 1 1 1 1 1 = + = + f u v f u u 1 1 1 3 = + = u= 60 u= 30 0 u u 0 u In the case where the image is real, u = 30 cm [First look at the case where the image is real.] 1 1 1 1 1 1 = = [Then look at the case where the image is f u v f u u virtual.] 1 1 1 1 = = u= 0 u= 0 u u 0 u In the case where the image is virtual, u = cm. Q5 u+ v= 3 v= 3 u [In this case u and v are object and image distances; it should be clear from the diagram we can relate them in this way.] 1 = = + f u v 7.5 u 3 u [We know that the image is real since it is formed on a screen.] 1 3 u u u( 3 u) = + 7.5 = 7.5 u 3 u u 3 u 3 ( ) ( ) 3u u 7.5 = 3( 7.5) = 3uu u 3u + 40 = 0 3 ( u1)( u 0) = 0 u = 1 or 0 [Solve the quadratic by factorising or formula.] The two possible object distances are u = 1 cm or u = 0 cm. Exercise 13.3 Page 38
EXERCISE 13.4 COMPOUND LENSES 1 1 Q1 (i) P = = = 4 f 0.5 P = 4 m 1 1 1 (ii) P = = =.5 f 0.4 P =.5 m 1 [Converging lens implies focal length is positive, also convert f to SI units] [Diverging lens implies focal length is negative, also convert f to SI units] 1 1 1 Q (i) P= f = = = 0.1 f P Converging lens, f = cm (ii) P= 1 f = 1 = 1 = 0.15 f P 8 Diverging lens, f = 1.5 cm [Power is positive implies lens is converging] [Power is negative implies lens is diverging] 1 1 1 1 Q3 P= P1 + P = + = + = 9 f1 f 0. 0.5 P = 9 m 1 ( )( ) ff 0.3 0. 1 0.06 3 Q4 (i) f = = = = =0.6 f1 + f 0.3 0. 0.1 5 f = 60 cm (ii) Diverging lens 1 1 1 Q5 (i) P= f = = = 0. f P 5 ( 15) ff f 1 1 f = 0 = f + f f + 15 1 1 0 f + 300 = 15f 5f =300 f =60 f = 60 cm (ii) Diverging lens 1 1 1 1 [Convert f 1 and f to SI units.] [The focal length of the compound lens is 11.11 cm.] [Focal length is negative implies lens is diverging] [The focal length of the compound lens is 0 cm.] [It does not matter whether you use SI units or not so long as you are consistent about the units you use.] [Focal length is negative implies lens is diverging] Exercise 13.4 Page 39
EXERCISE 14.1 VERNIER SCALE ON A SPECTROMETER Q1 Step A: The first line on the Vernier scale lines up a little bit past 40 30. Step B: The line on the Vernier scale that lines up best with a line on the main scale is 15. Step C: Combine these two readings to give a reading of 40 45. The reading is 40 45. Q Step A: The first line on the Vernier scale lines up a little bit past 133 00. Step B: The line on the Vernier scale that lines up best with a line on the main scale is 8. Step C: Combine these two readings to give a reading of 133 8. The reading is 133 8. Questions involving calculating angles using the Vernier scale on a spectrometer should not be asked in the Leaving Cert. examination. They are, however, of practical use, as a spectrometer may be used while carrying out experiments in the laboratory. Exercise 14.1 Page 40
EXERCISE 14. WAVELENGTH OF LIGHT 1 1 d = = N 700 3 3 Q1 [You can choose to put the answer in decimal, but this may lead to a rounding error in the answer. It is better to leave it as a fraction as decimals are only an approximation.] 1 3 7 nλ = d sinθ ( 1) λ = sin30 = 7.143 700 λ = 714 nm [You can give your answer as 7.14 7 m or you can use the prefix nano; n, 9 ).] Q (i) ( )( 9 6 n = d ) = ( ) λ sinθ 1 40.0 sinθ 9 40 1 1 sinθ = θ = sin = 6 1.1.0 0 θ = 1.1 9 6 (ii) n = d ( )( ) = ( ) λ sinθ 40.0 sinθ 9 840 1 4 sinθ = θ = sin = 6 4.8346.0 0 θ = 4.83 9 6 (iii) n = d ( )( ) = ( ) λ sinθ 3 40.0 sinθ 9 160 1 63 sinθ = θ = sin = 6 39.050.0 0 θ = 39.05 1 1 d = = = N 500 opposite 0.7 1 tanθ = = θ = tan ( 0.36) = 19.8 [Use trigonometry to get the value of θ.] adjacent 6 7 nλ = d sinθ 1 λ = sin19.8 = 6.774 3 3 6 Q3 λ = 677 nm ( ) ( ) d Q4 sinθ 1 nλ d n λ 6. n n 3.4 9 650 There are 3 images to the left of the zero order, 3 to the right, plus the zero order itself. There are a maximum of 7 bright fringes. Exercise 14. Page 41
EXERCISE 15.1 COULOMB S LAW Q1 Q 1 QQ F = 4πε d 1 6 6 ( )( 11 1 6 6 ) 3.6 1 14 ( ) ( 0.04) ( ) [Coulomb s law] F = = = 01.179 4π 8.9 5.696 π F = 01.18 N 1 QQ F = 4πε d 1 6 6 ( )( 11 1 8 ) 8 1 14 ( ) ( 0.03) ( ) [Coulomb s law, assume the permittivity of air is the same as the permittivity of free space.] F = = = 794.781 4π 8.9 3.04 π F = 794.78 N Q3 ε εε r ( ) Q4 = 0 = 7 8.9 = 6.3 ε = 6.3 11 F m 1 1 11 11 ε.6 60 ε= εε r 0 εr = = = = 1.91 ε 0 8.9 89 ε r =.9 1 11 Q5 ε= εε r 0 =.35( 8.9 ) =.0915 6 6 ( )( 1 0 ).0 F 11 16 4π (.0915 ) ( 0.003) ( 7.594 ) [It makes no difference to the size of the force if the charges are of opposite charge (positive and negative) or if they are the same. The force will be attractive if the charges are opposite and repulsive if they are like charges.] [Relative permittivity is not mentioned amongst the topics to be covered as described in the syllabus and to date (0) has not appeared in any Leaving Cert. exam paper. However, relative permittivity, ε r, is mentioned amongst the notation that is to be covered as part of the syllabus. It is covered in this textbook so that students will be prepared in the unlikely event that it comes up in a Leaving Cert. examination.] [Coulomb s law] = = = 84 551. π F = 84.55 kn [Answer can be given in kilonewtons or as 8.455 4 N.] Exercise 15.1 Page 4
Q6 [Look at the forces on A due to B and C separately, and then add them.] 6 6 ( )( 5 ) 1 ( ) ( ) 1 QQ 1 1 F = FAB = 4πε d 4π 8.9 0.04 F AB 11 5.0 = = 79.415 14 ( 5.686 )π 6 6 ( )( 8 ) 1 ( ) ( ) 1 QQ 1 1 F = FAC = 4πε d 4π 8.9 0.11 F AC 11 8.0 = = 59.116 13 ( 4.3076 )π [Electrostatic force on A due to B] [F AB = 79.415 N to the left] [Electrostatic force on A due to C] [F AC = 59.116 N to the right] F = F F = 79.415 59.116 = 0.99 [F AB and F AC act in opposite directions, A AB AC F A = 0.3 N to the left therefore subtract the forces and use the sign (direction) of the bigger one.] [Look at the forces on A due to B and C separately, and then add them.] FBA = F AB [As a result of Newton s third law you should note that the forces are equal in size and opposite in direction.] F BA =79.415 [F BA = 79.415 N to the right] 6 6 ( 5 )( 8 ) 1 ( ) ( ) 1 QQ 1 1 F = FAC = 4πε d 4π 8.9 0.07 [Electrostatic force on B due to C] 11 4.0 F AC = = 7.990 13 ( 1.7444 )π [F BC = 7.990 N to the right] FB = FBA + F BC = 79.415 + 7.990 = 35.405 [F BA and F BC act in the same directions, therefore just add the forces] F B = 35.41 N to the right [Look at the forces on A due to B and C separately, and then add them.] FCA = FAC and FCB =F BC [As a result of Newton s third law] FCA = 59.116 and F CB = 7.990 [Both forces act to the left.] FC = FCA + F CB = 59.116 + 7.990 = 13.6 [F CA and F CB act in the same directions, therefore just add the forces.] F C = 13.11 N to the left Exercise 15.1 Page 43
Q7 (i) (ii) 6 6 ( 3 )( 3 ) 1 ( ) ( ) 1 QQ 1 1 F = F = 4πε d 4π 8.9 0. 1 9.0 F = = 8.047 13 ( 3.56 )π F = 8.05 N Gm1m F = F = d 11 ( 6.7 )( 0.05)( 0.05) ( 0.) 13 1.675 F = = 1.675 0.01 F = 1.675 11 N (iii) The direction of the forces 11 [This is the electrostatic force.] [This is the gravitational force.] [The electrostatic force is repulsive, while the gravitational force is attractive.] Exercise 15.1 Page 44
EXERCISE 15. ELECTRIC FIELDS Q1 Q Q3 (i) Q4 (i) E = F Q 3 1 F = = 333 6 6 3 E = 333.33 N C 1 [Definition of electric field strength] = 1 Q E [Formula for electric field strength] 4πε d 19 19 1 1.6 1.6 9 F = = = 1.4306 1 π 9 9 4 8.9 1 3.56 π ( ) ( ) ( ) E = 1.43 9 N C 1 = 1 Q E [Formula for electric field strength] 4πε d 6 6 1 15 15 7 F = = = 1.341 1 13 4π 8.9 0.1 3.56 π ( ) ( ) ( ) E = 1.34 7 N C 1 F = = = 1.34 5 = 67.06 Q F = 67.06 N 7 6 (ii) E F EQ ( )( ) (ii) E A E A 1 Q 1 8 = = 4πε d 4πε 0.008 6 ( ) 1 15 ( ) ( 0.008) ( ) [Coulomb s law] 6 6 1 8 8 = = = 1.11766 4π 8.9.784 π E A = 1.1 9 N C 1 directed to the right E B E B 1 14 ( ) ( 0.037) ( ) [Electric field strength at X due to A] 9 [Because the charge at A is positive, the electric field strength is directed away from it.] 6 1 Q 1 5 = = [Electric field strength at X due to B, the 4πε d 4πε ( 0.037) distance is 45 mm 8 mm] 6 6 1 5 5 8 = = = 1.638 4π 8.9 4.87364 π E B = 1.638 8 N C 1 directed to the right [Because the charge at B is negative, the electric field strength is directed towards it.] E = 1.11766 9 + 1.1638 8 = 1.809 9 [Add to the vectors because they act in opposite directions.] E = 1.8 9 N C 1 directed to the right Exercise 15. Page 45
EXERCISE 16.1 POTENTIAL DIFFERENCE W 1 Q1 V = = = 4 Q 3 V = 4 V [Definition of potential difference] Q 18 W 8 V = = = 50 19 Q 1.6 V = 50 V W = Q = = 5 5 =.5 W = 0.05 J Q3 V W VQ ( )( ) 3 6 Exercise 16.1 Page 46
EXERCISE 16. CAPACITANCE Q = V = = 6 4 = 144 Q = 144 C Q1 C Q CV ( )( ) ( )( ) Q 1 4 ε A 8.9 0 C = = d 3 = 1.5 C = 1.5 11 F ( )( ) 11 1 4 ε A 8.9 0 Q3 C = = = 5.933 d 3 C = 5.93 13 F 13 [Definition of capacitance] [Definition of capacitance] [Capacitance of a parallel plate capacitor: note that y cm = y 4 m. Refer to p.7 for explanation as to why 1 m = 000 cm.] 1 1 W = CV = 00 1 = 0.0144 [Energy stored in a capacitor] W = 0.0144 J 6 Q4 ( )( ) ( )( ) 11 6 ε A 6.3 4 Q5 (i) C = = =.49 3 d 0.1 C =.49 1 F (ii) C Q CV ( )( ) 1 Q = = =.49 1 =.99 V Q =.99 11 C (iii) ( )( ) 1 11 [Capacitance of a parallel plate capacitor, note that y mm = y 6 m ] [Definition of capacitance] 1 1 1 W = CV =.49 1 = 1.794 [Energy stored in a capacitor] W = 1.794 J Exercise 16. Page 47
EXERCISE 17.1 JOULE S LAW AND KILOWATT HOURS Q1 I ( ) ( )( ) Q W = Rt = 6 5 10 = 1600 [Joule s law, change time to SI units.] W = 1.6 kj 3 W 3.6 4 W = I Rt R = = = I t 5 R = 0.8 Ω ( 5) ( 180) Q3 θ ( )( )( ) [It s equally correct to give your answer as 1 600 J or.16 4 J.] [Joule s law, change time to SI units.] Q = mc = 0.08 4180 9.3 = 39.9 [Using hint; θ = 4.8 15.5 ] = I = W I = 39.9 W Rt R = 9.1 t R = 9.1 Ω P Q4 P= IV I = = 60 V 30 ( 1.5) ( 150) [Q = heat produce = W] [This definition of power is based on combining Joule s law and Ohm s law. Ohm s law is covered in Ch.18; the question is therefore out of sequence and should not be asked in this chapter.] P 60 P= I R R = = = 881 [Joule s law, change time to SI units.] I 6 3 3 R = 881.67 Ω 4 1 = 0 00 = 3 = 18 00 = 3.6 W = W1 W = 3.6 = 1.64 W = 16.4 kw h Q5 W ( )( ) W ( )( ) Q6 P W Pt ( )( ) 4 3 4 [Energy = power time; 0 W bulb] [Energy = power time; 18 W bulb] W = t = = 900 80 = 7 000 J [The toaster uses 7 kj of energy each time.] 7 W = 7 000 365 =.68 J [Total energy use, based on 365 days per year] 7 4.68 J =.68 kj [Converting joules to kilojoules] 680 W = = 7.3 3600 [Converting kilojoules to kilowatt hours] W = 7.3 kw h Exercise 17.1 Page 48
EXERCISE 18.1 RESISTANCE 1 1 1 1 1 1 Q1 (i) = + = + = R = 16 R R R 0 80 16 1 R = 16 Ω V 3 (ii) V = IR I = = = 0.1875 R 16 I = 0.1875 A V 3 (iii) V = I0ΩR0Ω I0Ω = = = 0.15 R 0 I Ω = 0 0.15 A Alternative approach 0Ω [Resistors in parallel] [Ohm s law] [Using Ohm s law] = 4 4 = 0Ω 0.1875 0.15 5 5 (iii) I I= ( ) I Ω = 0 0.15 A [Current splits in the inverse ratio to the resistances of the resistors.] Q (i) R= R1+ R = 0 + 300 = 400 [Resistors in series] R = 400 Ω V 1 (ii) V = IR I = = = 0.03 R 400 I = 0.03 A V R [Using Ohm s law] (iii) I ( )( ) V 300Ω = 300Ω = 0.03 300 = 9 = 300Ω 9 V Alternative approach = 3 = 3 V300Ω V 1 4 4 = 9 [Voltage drops in ratio to the resistances of the resistors.] V = 300Ω 9 V (iii) ( ) (iv) The potential difference is reduced. [Because the ratio of resistances of the fixed resistor to variable resistor changes, a greater percentage of the total voltage (1 V) drops across the 0 Ω resistor.] Exercise 18.1 Page 49
V 6 Q3 (i) V = I1R1 I1 = = = 0.03 R 1 00 I = 0.03 A V 6 V = IR I = = = 0.06 R 0 I = 0.06 A V 6 V = I3R3 I3 = = = 0.0 R 3 300 I = 0.0 A (ii) There is no change. [Based on Ohm s law, the current through each resistor depends only on the potential difference across it and its resistance, neither of which is affected by varying a parallel resistor. The total current through the circuit is reduced because less current flows through the variable resistor, but the current through the fixed resistors is unaffected.] 1 1 1 1 1 9 4 000 Q4 (i) = + = + = RP = [Resistance of the parallel part of the circuit] RP R R 3 00 800 4 000 9 4 000 5350 R= R1 + R P = 150 + = = 594.44 9 9 R = 594.44 Ω 4 000 80 960 VP = V = = [Calculate the voltage across the parallel 5350 7 7 part of the circuit.] 960 VP 7 VP = IR800Ω I800Ω = = [Using Ohm s law] R (ii) ( 1) 800Ω 800 I = 6 800Ω A 535 Alternative approach V 1 54 (ii) V = IR I = = = R 675 5 350 9 00 5 54 6 I800Ω = I= = 1800 9 675 535 I Ω = 800 6 A 535 (iii) The potential difference increased. [Calculate the total current flowing in the circuit.] [Current splits in the inverse ratio to the resistances of the resistors.] [Lowering the resistance of the variable resistor causes the resistance of the parallel part of the circuit to be reduced. This results Exercise 18.1 Page 50
in a greater percentage of the total voltage (1 V) dropping across the 150 Ω resistor.] Exercise 18.1 Page 51
EXERCISE 18. RESISTIVITY Q1 Q Q3 ( ) π ( ) ( ) ( ) 3 7 Rπ d 4.8 0.3 4.3 π ρ = = = = 1.696 4l 4 0 80 ρ = 1.7 8 Ω m ( ) π ( ) ( ) ( ) 6 7 Rπ d 90 46 6.1364 π ρ = = = = 9.0934 4l 4 0.53.1 ρ = 9.1 7 Ω m 6 ( )( ) RA ρl 1.1 0.8 4 ρ = R = = = 4 8 l A 3.6 9 R = 4.44 Ω 8 [Resistivity formula] 7 Exercise 18. Page 5
EXERCISE 18.3 VERNIER SCALE ON A MICROMETER Q1 Step A: The line on the main scale that is directly in front of the Vernier scale is 5.5. Step B: The line on the Vernier scale that lines up with the centre line of the main scale is 1. Step C: Combine these two readings to give a reading of 5.6. Step D: There is no zero error, so there is nothing to subtract. The reading is 5.6 mm. Q Step A: The line on the main scale that is directly in front of the Vernier scale is 5.5. Step B: The line on the Vernier scale that lines up with the centre line of the main scale is 34. Step C: Combine these two readings to give a reading of 5.84. Step D: Subtract the zero error of 0.0 mm: 5.84 0.0 = 5.8. The reading is 5.8 mm. Questions involving calculating distance using the Vernier scale on a micrometer should not be asked in the Leaving Cert. examination. They are, however, of practical use, as a micrometer may be used while carrying out experiments in the laboratory. Exercise 18.3 Page 53
EXERCISE 18.4 WHEATSTONE BRIDGE Q1 Q R1 R3 50 35 35 50 = = R4 = = 175 R R 50 R 50 4 4 R 4 = 175 Ω R1 l1 8 68.1 8 31.9 = = R = = 3.747 R l R 31.9 68.1 R 4 = 3.75 Ω [Wheatstone bridge formula] [Metre bridge formula] 1 1 V1 = V = 3 3 3 = 1 [The potential difference across R 1 is 1 V.] = 3 1= [The potential at B equals the potential at A Q3 (i) ( ) V B V B = V (ii) ( 3) minus the potential difference across R 1 ] 40 30 V3 = V = = [Calculate the voltage across R 3 in the 14 31 31 same way as part (i).] 30 1 V = 3 = [Same approach as part (i)] D 31 31 V D = 1 V 31 (iii) Current flows from D to B. (iv) R1 R3 0 40 40 00 = = R4 = = 80 R R 00 R 0 4 4 R 4 = 80 Ω [Because D is at a higher potential] [Galvanometer is zero means that the Wheatstone bridge is balanced and the formula can be applied] Exercise 18.4 Page 54
EXERCISE 0.1 CURRENT IN A MAGNETIC FIELD Q1 (i) Il ( )( )( ) F = B = 0..5 = 1 [Force on a current carrying conductor] F = 1 N (ii) Down Q (i) Il ( )( )( ) [Use Fleming s left hand rule] F = B =.5 0.1 6 = 1.5 [It is only the cm long sides that F = 1.5 N (ii) ( )( ) experience a force due to the magnetic field, as the other sides are parallel to the field.] T = Fd = 1.5 0.05 = 0.075 [Torque due to a couple] T = 0.075 N m Q3 ( 6 )( 3 )( ) F = qvb = 3 4 5 = 0.06 [Force on a charged particle] F = 0.06 N Q4 (i) F qvb ( )( )( ) (ii) = = 1.6 1.6 50 = 6.4 F = 6.4 11 N 19 6 11 11 5 ( 6.4 )( 6.68 ) 6 ( 1.6 ) mv Fr F = m= = = 1.67 r v m = 1.67 7 kg 7 [Centripetal force] Exercise 0.1 Page 55
EXERCISE 0. ELECTROMAGNETIC INDUCTION Φ = = 3.5 0. = 0.7 Q1 BA ( )( ) Φ = 0.7 Wb Q π π( 0.5) Q3 [Magnetic flux] A= r = π = 16 [Area of a circle: radius equals half the diameter] Φ 16 48 Φ = BA B = = ( 3) = = 15.8 A π π B = 15.8 T Φfinal Φinitial 6 E = = = 0 t 0. E = 0 V 36π = = = 5 36π A = m 5 Q4 (i) A πr π( 1.) 36π 8π = = = 5 5 π Φ = 8 Wb 5 (ii) Φ BA ( 3) s s.4 (iii) v = t = = = 0.4 t v 6 t = 0.4 s Φ Φ 8π final initial 5 0 54π (iv) E = = = = 33.99 t 0.4 5 E = 33.9 V Q5 (i) BA ( )( ) Φ 1 = = 6.8 0.1 = 0.0979 Φ 1 = 0.0979 Wb (ii) Φ = 0 Wb 1 1 (iii) T = = = 0.05 f 0 T 1 t = = = 0.015 4 80 t = 0.015 s dφ 0.0979 E = N = 400 =3133.44 dt 0.015 E = 31 334.4 V (iv) ( ) [Faraday s law] [The minus can be ignored as we are only interested in the magnitude of the induced emf.] [Velocity formula] [Area of the square = 1 cm 1 cm] [When the loop is aligned such that the plane of the square is parallel to the field] [Periodic time: one complete revolution] [Flux changes from max to zero every quarter revolution] [Faraday s law, using calculus notation] Exercise 0. Page 56
EXERCISE 0.3 TRANSFORMERS V0 16 Q1 V rms = = = 8 = 11.3137 = 11.31 V Q Q3 Q4 V rms [Root mean squared voltage] I0 Irms = I0 = Irms = = 14.14 [Root mean squared current] I 0 = 14.14 A V N i p 30 400 30 0 = = Vo = = 57.5 Vo Ns V o 0 400 V o = 57.5 V V N i p 000 000 000 50 = = Ns = = 50 Vo Ns 50 N s 000 N s = 50 [Transformer formula] Exercise 0.3 Page 57
EXERCISE 1.1 CATHODE RAY TUBES Q1 (i) ( )( ) 19 17 10 ev = 10 1.6 J = 1.9 J [1 ev = 1.6 19 J] E = 1.9 17 J 1 1 (ii) = 1.9 = ( 9.1 ) 17 31 E mv v [Kinetic energy of the electron] 17 ( ) 1.9 v = = 4.1978 = 6.5 31 9.1 v = 6.5 6 m s 1 13 6 1 1 qv mv v Q ( = 1.6 )( 1 ) = ( 9.1 ) 19 3 31 ( 19 3 )( ) 1.6 1 v = = 4.1978 = 6.5 31 9.1 v = 6.5 7 m s 1 Q3 ( )( )( ) 15 7 F = qvb = 1.6 19. 7 8 =.816 11 [Force on a moving charge] F =.8 11 N Q4 (i) E qv ( )( ) = = 1.6 = 3. E = 3. 16 J 19 3 16 1 1 E mv v (ii) = 3. = ( 9.1 ) 16 31 16 ( ) 3. v = = 7.033 =.65 31 9.1 v =.65 7 m s 1 (iii) F qvb ( )( )( ) 14 7 = = 1.6.65 5 =.1 F =.1 14 N 19 7 3 14 ( )( ) 31 7 mv mv 9.1.65 (iv) F = r = = = 0.0301 14 r F.1 r = 3.01 cm Exercise 1.1 Page 58
EXERCISE 1. PHOTOELECTRIC EFFECT Q1 8 c 3.0 16 c = fλ f = = = 1.5 [Velocity of a wave formula] 9 λ 0 34 16 18 E = hf = 6.6 1.5 = 9.9 [Energy of a photon] ( )( ) 18 18 9.9 E = 9.9 J = ev = 61.875 ev [1 ev = 1.6 19 J] 19 1.6 E = 61.875 ev Q (i) ( )( ) (ii) E = qv = 1.6 19 50 3 = 8 15 [Energy of a charged particle in an E = 8 15 J 15 E 8 E = hf f = = = 1.1 34 h 6.6 f = 1.1 19 Hz Q3 (i) Φ hf ( )( ) Q4 Q5 (ii) = 0 = 6.6 1.08 = 7.18 Φ = 7.18 19 J 34 15 19 19 electric field] [Work function] 1 hf = Φ + mv max [Einstein s photoelectric law] 34 15 19 1 31 ( 6.6 )( 1.5 ) = 7.18 + ( 9.1 ) vmax 19 19 31 9.9 7.18 = 4.55 v max.77 v v 4.55 V max = 7.8 5 m s 1 ( ) 19 11 5 max = = = 31 max 6.093 7.8 8 c 3.0 14 c = fλ f = = = 6.667 [Velocity of a wave formula] 9 λ 450 34 14 19 E = hf = 6.6 6.667 = 4.4 [Energy of a photon] ( )( ) hf = 4.4 19 J > Φ = 3.6 19 J, therefore we can conclude that photoemission can occur. 1 hf = Φ + mv max [Einstein s photoelectric law] 1 19 19 1 31 hf = Φ + mvmax 4.4 = 3.6 + ( 9.1 ) v max 19 19 4.4 3.6 11 5 vmax = v = = 31 max 1.758 4. 4.55 V max = 4. 5 m s 1 1 hf = Φ + mv max [Einstein s photoelectric law] 6.6 1.1 = + 3.5 = 7.6 3.5 = 3.76 ( )( ) Φ Φ 34 15 19 19 19 19 19 Φ 3.76 Φ = hf0 f0 = = = 5.7 34 h 6.6 f 0 = 5.7 14 Hz 14 [Work function] Exercise 1. Page 59