Answer, Key Homework 0 David McIntyre 453 May 0, 004 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. Chapter 9 problems. 00 (part of ) 0 points An object of mass m is moving with speed v 0 to the right on a horizontal frictionless surface, as shown, when it explodes into two pieces. Subsequently, one piece of mass 5 m moves with a speed v /5 v 0 to the left. v 0 m The speed v 3/5 of of the other piece the object is. v 3/5 v 0.. v 3/5 v 0 3. 3. v 3/5 7 v 0 5. 4. v 3/5 3 v 0. 5. v 3/5 v 0. correct The horizontal component of the momentum is conserved, so 0 + m v 0 5 m v /5 + 3 5 m v 3/5 0 + m v 0 ( 5 m v ) 0 + 3 5 m v 3/5 mv 0 5 m v 0 + 3 5 m v 3/5 3 5 v 3/5 6 5 v 0 v 3/5 v 0. 00 (part of ) 0 points An open train car moves with speed 9 m/s on a flat frictionless railroad track, with no engine pulling the car. It begins to rain. The rain falls straight down and begins to fill the train car. The speed of the car. stays the same.. decreases. correct 3. increases. Using Newton s second law, we have d P dt 0, since no external forces act on the train in the horizontal direction. With no rain, the train will move at a constant velocity; however, when it starts to rain, and the rain starts to fill the car, the mass of the train changes. Thus, m d v dt v d m dt. Since d m is positive; i.e., the mass of the dt train is increasing with accumulating rain, d v should be negative; i.e., the speed of the dt train should decrease. 003 (part of ) 5 points Consider the set up of a ballistic pendulum where.4 g, m 3.97 kg, h 0.43 m. The acceleration of gravity is 9.8 m/s. v m v f + m Find the final velocity of the system ( + m ) immediately after the collision and before the pendulum starts to swing upwards. Correct answer:.838 m/s. From the setup, the final velocity of the collision is the initial velocity of the subsequent h
Answer, Key Homework 0 David McIntyre 453 May 0, 004 swing. During the swinging process the total energy is conserved Therefore E i E f ( + m ) vf ( + m ) g h. v f g h (9.8 m/s ) (0.43 m).838 m/s. 004 (part of ) 5 points Find v, the initial speed of. Correct answer: 76.88 m/s. The linear momentum is conserved in a collision Therefore v + m p i p f v ( + m ) v f v f (0.04 kg) + (3.97 kg) (.838 m/s) (0.04 kg) 76.88 m/s. 005 (part of ) 0 points A 83 kg man holding a 0.336 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 5.7 m/s (relative to the ground) and then catches the ball after it rebounds from the wall. Hint: Ignore the projectile motion of the ball, and assume that it loses no energy in its collision with the wall. How fast is he moving after he catches the ball? Correct answer: 0.660 m/s. Let M and m be the mass of the man and the ball respectively. Let v be the speed of the ball. Using momentum conservation twice, we can get the speed the man moves after he catches the ball back. That is, when the man throws the ball away, M v m v and when the man catches the ball bounced back from the wall, (M + m) V M v + m v m v. Solving for V, we get, V m v M + m 006 (part of ) 0 points Note: Remember that the ball is thrown with a speed of 5.7 m/s relative to the ground each time it is thrown. How many times does the man have to go through this process before his speed reaches at least.7 m/s relative to the ground? Correct answer:. If he repeats this n times, he ll move at a speed V n m n v M + m. If he is to reach V, he should repeat this ( ) M + m n V m v times. 007 (part of 4) 3 points A massless spring with force constant k 44 N/m is fastened at its left end to a vertical wall, as shown below. The acceleration of gravity is 9.8 m/s. k M C M D C D Initially, block C (mass m C 3.57 kg) and block D (mass m D.86 kg) rest on a horizontal surface with block C in contact with the spring (but not compressing it) and with block D in contact with block C. Block C is then moved to the left, compressing the spring
Answer, Key Homework 0 David McIntyre 453 May 0, 004 3 a distance of 0.4 m, and held in place while block D remains at rest as shown in below. d C D µ Determine the elastic energy U stored in the compressed spring. Correct answer: 35.8 J. Should use the expression for the energy stored in a spring: U k x (44 N/m) (0.4 m) 35.8 J. 008 (part of 4) 3 points Block C is then released and accelerates to the right, toward block D. The surface is rough and the coefficient of friction between each block and the surface is µ 0.4. The two blocks collide instantaneously, stick together, and move to the right. Remember that the spring is not attached to block C. Find the speed v C of block C just before it collides with block D. Correct answer: 4.05759 m/s. Apply conservation of energy or workenergy theorem to this problem. Kinetic energy of block C is K m C v ; work done (or C energy dissipated by) friction is W f µ F n d. Therefore we have m C v U µ m C C g d. Solving for v C v C m C (U µ m C g d) 4.05759 m/s. Correct alternate solution Computing F net dx, where F net k x µ F n, to find the kinetic energy, and then computing the speed. 009 (part 3 of 4) points Find the speed v f of blocks C and D just after they collide. Correct answer:.58 m/s. In this configuration, we have the conservation of momentum: Solving for v f m C v C (m C ) v f. v f m C v C m C.58 m/s. 00 (part 4 of 4) points Find the horizontal distance the blocks move before coming to rest. Correct answer: 0.65054 m. The blocks come to rest when all their kinetic energy has been dissipated; i.e., KE Work done by frictional force (m C ) v f µ (m C ) g l Solving for l v f l µ g (.58 m/s) (0.4) (9.8 m/s ) 0.65054 m. Alternate Solution: F m a F a m µ (m C ) g (m C ) µ g 4.58 m/s v v0 + a l l v v 0 a 0.65054 m.
Answer, Key Homework 0 David McIntyre 453 May 0, 004 4 0 (part of ) 0 points A bullet of mass 5.5 g moving with an initial speed 35 m/s is fired into and passes through a block of mass.85 kg, as shown in the figure. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 400 N/m. v m v + k x m v i (0.0055 kg) (77.8 m/s) (400 N/m) (0.0 m) + (0.0055 kg) (35 m/s) (86.333 J) + (0.008 J) (7.869 J) 86.436 J E 86.436 J. If the block moves a distance. cm to the right after the bullet passed through it, find the speed v at which the bullet emerges from the block. Correct answer: 77.8 m/s. From conservation of energy Hence, M V 0 k x. 03 (part of ) 0 points Consider the general case of collisions of two masses m with m m along a frictionless horizontal surface. Denote the initial and the final center of mass momenta to be and p i cm p + p, p f cm p + p. And the initial and final kinetic energies to be k x V 0 0.65964 m/s. M Then from momentum conservation and K i K + K, K f K + K. m v 0 M V 0 + m v, or v v v v 0 M V 0 m (35 m/s) 77.8 m/s. (.85 kg) (0.65964 m/s) (5.5 g) (0.00 kg/g) 0 (part of ) 0 points Find the magnitude of the energy lost in the collision. Correct answer: 86.436 J. The energy lost is E K f + U f K i m For an elastic collision which pair of statements is correct?. p i cm < p f cm, K i < K f. p i cm > p f cm, K i K f 3. p i cm > p f cm, K i > K f 4. p i cm p f cm, K i > K f 5. p i cm > p f cm, K i < K f 6. p i cm p f cm, K i K f correct
Answer, Key Homework 0 David McIntyre 453 May 0, 004 5 7. p i cm < p f cm, K i K f 8. p i cm < p f cm, K i > K f 9. p i cm p f cm, K i < K f Momentum and energy are always conserved in elastic collision. 04 (part of ) 0 points Now consider an elastic head-on collision again with initial velocity of to be v and of m to be v 0. Find the final speed v f of m.. v f 3 v. v f v 3. v f 3 4 v 4. v f 7 v 5. v f 3 5 v 6. v f v 7. v f v 8. v f 5 v 9. v f 3 v correct 0. v f 4 v The CM velocity is given by V CM i m i v i v + m i m i as v 0. Thus, in the CM frame, m has a velocity v (CM) v V CM V CM v + m. After a one dimensional elastic collision, each body changes the direction of its momentum with the magnitude of the momentum conserved. Note: In the CM frame the total momentum is still zero if each velocity changes sign. Hence, the final velocity of m in the CM frame is v (CM),f V CM v + m The velocity v,f lab (vf ) in the lab frame is v lab,f v(cm),f + V CM V CM v + m 3 v. To sum up, we have the following general expression, v,f v cm v,i. For the present case,v,i 0, v cm v 3, so v,f 3 v 05 (part of 3) 0 points Two balls with masses and m are on the x-axis. Ball has an initial velocity v > 0 along the positive x-axis and ball m is initially at rest. The balls collide elastically and remain on the x-axis after the collision. If m, what is the final velocity v of the ball?. v +v. + v < v < 3. v v 4. v 0 correct 5. v + v Basic Concepts: Conservation of Energy, where v 0: v v + m v
Answer, Key Homework 0 David McIntyre 453 May 0, 004 6 Conservation of Momentum, where v 0: v v + m v Solving for v and v yields: ( v m m + m ( v m + m ) v () ) v () All solutions can be determined using the above equations () and (). Think of playing pool. You hit the cue ball with velocity v dead center on the 8 ball v 0. The cue ball stops and the 8 ball proceeds with the same velocity as the cue ball originally had. Part : Using eqn (), where m m, ( ) m m v v 0 m + m Part : Using eqn (), where m m, ( ) m v v +v m + m 06 (part of 3) 0 points If m, what is the final velocity v of the ball m?. v v. + v < v < 3. v + v 4. v 0 5. v +v correct 07 (part 3 of 3) 0 points In the limit, when m, what is the final velocity v of the ball?. v 0. v +v 3. + v < v < 4. v v correct 5. v + v Think of throwing a golf ball with velocity v against a concrete wall v 0. The golf ball hits the wall, bounces back (reverses direction) and has the same speed as it originally had (elastic collision). Part 3: Using eqn (), where 0, m ( m ) v m v v m + Part 4: Using eqn (), where 0, m ( ) v m v 0 m + 08 (part of ) 0 points A.5 g bullet is fired horizontally into a kg wooden block resting on a horizontal surface ( µ 0.57 ). The bullet goes through the block and comes out with a speed of 86 m/s. The acceleration of gravity is 9.8 m/s. If the block travels 5.64 m before coming to rest, what was the initial speed of the bullet? Correct answer: 4.893 m/s. Let m be the mass of the bullet and M the mass of the block. From momentum conservation, we have, mv mv + Mv From energy conservation, we have, Thus we have, Mv µmgd v µgd v v + M m v v + M m µgd
Answer, Key Homework 0 David McIntyre 453 May 0, 004 7 09 (part of ) 5 points A(n) 54 kg mass is sliding on a horizontal frictionless surface with a speed of m/s when it collides with a 87 kg mass initially at rest, as shown in the figure. The masses stick together and slide up a frictionless track at 60 from horizontal. The acceleration of gravity is 9.8 m/s. 60 87 kg m/s 54 kg 9.8 m/s What is the speed of the two blocks immediately after the collision? Correct answer: 9.4337 m/s. Let : v speed of blocks after collision 54 kg, m 87 kg, v m/s, and g 9.8 m/s. The initial momentum of block is v. When sticks with m, the resulting momentum p is equal to that initial momentum; thus, p ( + m ) v v. Therefore, the speed of the combined masses is v v + m (54 kg) ( m/s) (54 kg) + (87 kg) 9.4337 m/s. h 6 kg 9.8 m/s To what maximum height h above the horizontal surface will the masses slide? Correct answer: 4.54056 m. The kinetic energy K of the combined system is K ( + m ) v m ( + m ) v At the maximum height, all the kinetic energy is converted to potential energy, and so ( + m ) g h K Solving for h, we get h v g [ + m m ( + m ) v ] ( m/s) (9.8 m/s ) [ (54 kg) (54 kg) + (87 kg) 4.54056 m. 0 (part of ) 0 points A 30-06 caliber hunting rifle fires a bullet of mass 0.058 kg with a velocity of 84 m/s to the right. The rifle has a mass of 6.4 kg. What is the recoil speed of the rifle as the bullet leaves the rifle? Correct answer: 0.7903 m/s. From conservation of momentum p 0 0 p b + p r ] 00 (part of ) 5 points or 0 m b v b + m r v r.
Answer, Key Homework 0 David McIntyre 453 May 0, 004 8 Hence V m b v b m r (0.058 kg) (84 m/s) (6.4 kg) 0.7903 m/s. 0 (part of ) 0 points If the rifle is stopped by the hunter s shoulder in a distance of 3.88 cm, what is the magnitude of the average force exerted on the shoulder by the rifle? Correct answer: 4.589 N. The average force is The time of contact is F av p r t. t s v r 0.079 s, since the average velocity from the guns initial velocity (v v r ) to when it stops (v 0) is v v r assuming constant de-acceleration. Then the average force is F av m r v r t (6.4 kg) (0.7903 m/s) (0.079 s) 4.589 N. 03 (part of ) 5 points Given: G 6.6759 0 N m /kg Two hypothetical planets of masses 3.8 0 3 kg and 6.5 0 3 kg and radii 3.6 0 6 m and 5.9 0 6 m, respectively, are at rest when they are an infinite distance apart. Because of their gravitational attraction, they head toward each other on a collision course. When their center-to-center separation is 6.5 0 8 m, find their relative velocity. Correct answer: 459.459 m/s. At infinite separation the potential energy U is zero, and at rest the kinetic energy K is zero. Since energy is conserved we have 0 v + m v G m d The initial momentum is zero and momentum is conserved, so 0 v m v. Combine these two equations to find G v m d ( + m ) 6.5 0 3 kg (6.6759 0 N m /kg ) (6.5 0 8 m) (.03 0 4 kg) 89.95 m/s v G d ( + m ) 3.8 0 3 kg (6.6759 0 N m /kg ) (6.5 0 8 m) (.03 0 4 kg) 69.509 m/s. where + m (3.8 0 3 kg) + (6.5 0 3 kg).03 0 4 kg. The relative velocity is v r v ( v ) v + v 89.95 m/s + 69.509 m/s 459.459 m/s. 04 (part of ) 5 points Find the total kinetic energy of the planets just before they collide. Hint: Both energy and momentum are conserved. Correct answer:.73487 0 30 J. From conservation of energy the total kinetic energy of the system just before the.
Answer, Key Homework 0 David McIntyre 453 May 0, 004 9 planets collide is the opposite of the gravitational potential energy calculated for a distance d r + r : K tot G m r + r (6.6759 0 N m /kg ) (3.8 03 kg) (6.5 0 3 kg) (3.6 0 6 m) + (5.9 0 6 m).73487 0 30 J. For the minimum velocity case, the final kinetic energy of the rocket is zero. Since at the end the rocket will be infinitely far away from Planet-I, the final potential energy of the rocket is also zero. Conservation of energy implies K i + U i K f + U f 0. Thus K i 0 U i G M m R m g I R v i g R 05 (part of ) 0 points For this problem, which consists of parts, we assume that we are on Planet-I. The radius of this planet is R 490 km, the gravitational acceleration at the surface is g I.48 m/s, and the gravitational constant G 6.6759 0 N m /kg. The mass of Planet-I is not given. Not all the quantities given here will be used. Suppose a rocket of mass m 5900 kg is projected vertically upward from the surface of this planet. It stops at a point h 0935. km from the surface of the planet. Caution: Here the gravitational acceleration decreases as the rocket travels away from Planet-I. Determine the kinetic energy (in Joules) of the rocket immediately after it is fired off. Correct answer: 5.0959 0 0 J. For r > R, the potential energy of the rocket is U G M m. Since g I G M r R, U m g I R. Apply conservation of en- r ergy: E K i + U i K f + U f K i U f U i m g I R /r ( m g I R) mg I R( /n). 06 (part of ) 0 points Find the minimum initial velocity of the rocket such that it will escape the gravitational field of Planet-I. Note: You can work on this part even if you did not get Part. Correct answer: 46.85 m/s.