PY1052 Problem Set 6 Autumn 2004 Solutions

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1 PY052 Problem Set 6 Autumn 2004 Solutions () The mass of the Earth is kg and the mass of the Moon is kg. The distance between them is m, and the Earth s radius is R E = m. (a) How far is the centre of mass of the Earth Moon system from the centre of the Earth? (b) What is this distance in terms of Earth radii? We will great the Earth and the Moon as two points and take the zero of the axis joining their centres to be at the centre of the Earth. The centre of mass of the Earth Moon system is then given by X com = m EX E + m M X M m E + m M = ( kg)(0) + ( kg)( m) kg = m = 4647 km For comparison, the radius of the Earth is about 6370 km, so the centre of mass of the Earth Moon system is actually inside the Earth. (2) A slab with total length 22.0 cm, width 3.0 cm, and height 2.80 cm is made up half of iron (density = 7.85 g/cm 3 ) and half of aluminum (density = 2.70 g/cm 3 ). Where is the centre of mass of the slab? The easiest way to do this problem is to treat each of the two halves of the slab as a point having the total mass of that substance and located at the centre of mass of each half. Because the shape of each half-slab is regular, we know that the centre of mass of each half will be located right in the middle of that half. We will use a set of coordinate axes with its origin right at the centre of the whole slab, with the X axis measuring distance along the length of the slab, so that all the aluminum half will have positive X values and all the iron half will have negative X values. We can find the total mass of iron and aluminum using the corresponding densities and the volume of each half of the slab: X alum = +.0 cm/2 = cm X iron =.0 cm/2 = 5.50 cm V half =.0 cm 3.0 cm 2.80 cm = 400 cm 3 m iron = ρ iron V iron = (7.85 g/cm 3 )(400 cm 3 ) m iron = gm m alum = ρ alum V alum = (2.70 g/cm 3 )(400 cm 3 ) m alum = gm

2 We can now find the centre of mass as we did in the previous problem: X com = m ironx iron + m alum X alum m iron + m alum = ( g)( 5.50 cm) + ( g)(+5.50 cm) g = 2.68 cm The centre of mass is 2.68 cm from the centre of the slab; the minus sign reflects the fact that the centre of mass is on the negative X axis, in the iron half of the slab, as we expected. (3) A railroad flatcar of mass M =, 200 kg can roll without friction along a straight horizontal track. Initially, a man of mass m = 80 kg is standing on the car, which is moving to the right with speed v 0 = 4.50 m/s. What is the change in the velocity of the car if the man runs to the left so that his speed relative to the car is v rel = 3.0 m/s? The total momentum of the man + the flatcar will be conserved, because the rails are frictionless and so there are no external forces. If the final velocity of the man is v f and the final velocity of the flatcar is V f, then mv 0 + Mv 0 = mv f + MV f Here, v f and V f are as seen by an observer watching the flatcar go by. We are given the new speed of the man relative to the flatcar. Since the flatcar is moving to the right and the man runs to the left, the man s speed as seen by someone watching the flatcar go by will be v f = V f 3.0 m/s. Thus, (m + M)v 0 = m(v f 3.0 m/s) + MV f (m + M)v 0 = (m + M)V f m(3.0 m/s) V f = v 0 + m (3.0 m/s) M + m = 4.50 m/s + = 4.69 m/s 80 kg (3.0 m/s) 280 kg Because the running man s momentum is to the left, the flatcar had to increase its speed to the right in order to keep the total momentum of the man+flatcar constant. (4) A 9 kg man lying on a surface with negligible friction shoves a 68 kg stone away from him at a speed of 4.0 m/s. What velocity does he acquire as a result? The total momentum of the man + stone will be conserved, because there are no external forces. Thus, p i = p f 0 = (9 kg)v m + (68 kg)(4.0 m/s) v m = 2.99 kg m/s The negative sign indicates that the man moves in the opposite direction as the stone. (5) Cian, whose mass is 80 kg, and Eilish, who is lighter, are enjoying a peaceful ride on a lake in a 30 kg canoe, seated on two seats that are 3.0 m apart and located symmetrically with respect to the center of the canoe. When the canoe is at rest in the water, they exchange seats, and notice that the canoe has moved 40 cm relative to a submerged log. What is Eilish s mass?

3 M m X Before switch 3.0 m 3.0 m m M X After switch m = 3.4 m m = 2.6 m The key here is that the centre of mass of Cian+Eilish+the canoe does not change when they switch places. The diagram shows the situation before and after they switch. The dotted vertical line marks the location from which we will measure distances, which is the initial location of the centre of the canoe. The X marks the centre of mass since Cian is heavier than Eilish, it is located on his side of the canoe. We can find Eilish s mass by writing expressions for the centre of mass before and after they switch and setting them equal; writing Cian s mass M, Eilish s mass m and the mass of the canoe m c, we have X = M( 3.0 m) + m c (0) + m(+3.0 m) Before the switch X = m( 3.4 m) + m c ( 0.4) + M(+2.6 m) After the switch 3M + 3m = 3.4m 0.4m c + 2.6M 6.4m = 5.6M 0.4m c m = 5.6M 0.4m c (80 kg) 0.4(30 kg) = 6.4 = 68 kg (6) A radioactive nucleus of mass kg moving with a speed of v i = m/s divides into two pieces, one of mass m = kg and the other of mass m 2 = kg. The lighter particle moves along a path making an angle of 25 with the original direction of motion of the nucleus, while the path of the heavier particle makes an angle of 70 with the original direction of motion. (a) What are the speeds of each of the two particles? (b) What is the change in the kinetic energy during this decay process? 70 deg 25 deg

4 (a) The momentum is conserved because no external forces act. The momentum in the x and y directions are conserved separately this gives us two momentumconservation equations which must both be satisfied: p x,i = p x,f (m + m 2 )v i = m v x,f + m 2 v 2x,f (m + m 2 )v i = m v,f cos(25) + m 2 v 2,f cos(70) () p y,i = p y,f 0 = m v y,f + m 2 v 2y,f 0 = m v,f sin(25) m 2 v 2,f sin(70) (2) The only two unknowns in the two marked equations are v,f and v 2,f, and so we can solve for them both: (m + m 2 )v i = m v,f cos(25) + m 2 v 2,f cos(70) () 0 = m v,f sin(25) m 2 v 2,f sin(70) (2) m v,f = m 2v 2,f sin(70) sin(25) (m + m 2 )v i = [ m 2v 2,f sin(70) ] cos(25) + m 2 v 2,f cos(70) sin(25) () 4.00v i = 4.63v 2,f v 2,f = 5.48v 2,f v 2,f = 0.73v i = m/s Plugging this value for v 2,f into either () or (2) let s us solve for v,f : 0 = m v,f sin(25) m 2 v 2,f sin(70) (2) m v,f sin(25) = m 2 v 2,f sin(70) v,f = m 2v 2,f sin(70) m sin(25) = 2.3( m/s) sin(70).7 sin(25) v,f = m/s (b) The change in energy is given by KE = KE f KE i = [ 2 m v 2,f + 2 m 2v 2 2,f] 2 (m + m 2 )v 2 i = = 2 [( kg)( m/s) kg)( m/s) kg)( m/s) 2 ] = 2 [ J J J] = KE = J The kinetic energy has increased due to the decay potential energy that held the nucleus together has been transformed into kinetic energy of the particles after the decay.

5 (7) A.0 kg block at rest on a horizontal frictionless surface is connected to an unstretched spring with spring constant k = 200 N/m whose other end is fixed to a wall. A 2.0 kg block moving at 4.0 m/s collides with the.0 kg block. If the collision is completely inelastic (the blocks stick together), what is the maximum compression of the spring? Regardless of whether the collision is elastic or inelastic, the momentum will be conserved. Therefore, we can find the speed of the two blocks stuck together just after their collision: p i = p f (2.0 kg)(4.0 m/s) + (.0 kg)(0) = (2.0 kg +.0 kg)v f v f = 2.67 m/s After the collision, the total energy will be conserved. The total energy is made up of the kinetic energy of the blocks and the potential energy of the spring. Taking the initial time to be just after the collision and the final time to be when the spring is maximally compressed, KE i + PE i = KE f + PE f 2 mv2 i + 2 kx2 i = 2 mv2 f + 2 kx2 f 2 (3.0 kg)(2.67 m/s)2 + 0 = 0 + (200 N/m)x2 2 x = m = 32.7 cm (8) A cart with a mass of 340 g moving on a frictionless linear air track at an initial speed of.2 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in the same direction at 0.66 m/s. (a) What is the mass of the second cart? (b) What is its speed after the collision? In this situation, we cannot find both the mass of the second cart and its speed after the collision only by considering conservation of momentum we have two unknowns, and the conservation of momentum only gives us one equation. However, since the collision is elastic, we know that the kinetic energy is also conserved. This means that p i = p f m v i + m 2 = m v f + m 2 v 2f KE i = KE f 2 m vi m 2v2i 2 = 2 m vf m 2v2f 2 The final speeds of the two carts in terms of their masses and initial speeds are found by simultaneously solving these two equations, and are given by the relations v f = m m 2 v i + 2m 2 m + m 2 m + m 2 2m v 2f = v i + m 2 m m + m 2 m + m 2 In the first relation, we know everything except for m 2, so we can use this equation to find the mass of the second cart. We have since = 0:

6 v f = m m 2 v i m + m 2 (m + m 2 )v f = (m m 2 )v i m 2 (v f + v i ) = m (v i v f ) v i v f m 2 = m v f + v i m 2 = (0.340 kg) m 2 = kg = 98.7 gm We can now find v 2f from the second relation above: v 2f = 2m v i + m 2 m m + m 2 m + m 2 v 2f = 2m v i m + m 2 since = 0 v 2f = 2(0.340 kg) (.20 m/s) ( ) kg v 2f =.86 m/s (b) A 4.5-kg dog stands on an 8-kg flatboat 6. m from the shore. He walks 2.4 m along the boat toward the shore and then stops. Assuming there is no friction between the boat and the water, how far is the dog from the shore after he moves? 6. m xboat,i x dog,f xboat,f The key here is that the centre of mass of the dog+boat will not move when the dog walks along the boat toward the shore. The reason the dog has not actually gotten 2.4 m closer to the shore is that the boat has moved away from the shore a distance that keeps the centre of mass of the dog+boat stationary. If the distance the boat has moved away from the shore is x boat, then the total distance moved by the dog toward the shore is 2.4 m x boat and the final distance of the dog from the shore is x dog,i 2.4 m + x boat where x dog,i = 6. m. Setting the initial centre of mass of the dog+boat equal to the final centre of mass, measuring positions from the shore:

7 M boat x boat,i + M dog x dog,i = M boat x boat,f + M dog x dog,f M boat x boat,i + M dog x dog,i = M boat [x boat,i + x boat ] + M dog [x dog,i 2.4 m + x boat ] 0 = M boat x boat + M dog ( x boat 2.4 m) x boat = M dog (2.4 m) M dog + M boat x boat = 4.5 kg (2.4 m) = 0.48 m 4.5 kg kg Since the boat moves away from the shore 0.48 m, the final distance of the dog from the shore is x dog,i 2.4 m + x boat = 6. m 2.4 m m = 4.8 m (2b) A railroad freight car of mass M = kg collides with a stationary smaller railroad car. They couple together and continue along the track, with 27% of the initial kinetic energy being transformed into thermal energy, sound, vibrations, etc. What is the mass of the smaller car? Momentum will be conserved in the collision; this enables us to write the final speed in terms of the initial speed: p i = p f MV i + 0 = (m + M)V f V f = M m + M V i Now use the other information given the final kinetic energy is 73% of the initial kinetic energy, substituting in for V f in terms of V i : KE f = 0.73KE i 2 (m + M)Vf 2 = 2 MV i 2 (0.73) 2 (m + M)[ M m + M V i] 2 = 2 MV i 2 (0.73) M m + M = m = 0.27M m = 0.37M = kg (3b) Block of mass m is at rest on a long, frictionless table that is up against a wall. Block 2 of mass m 2 is placed between block and the wall and sent sliding to the left, toward block with a constant speed of. Assuming that all collisions are elastic, find the value of m 2 (in terms of m ) for which both blocks move with the same velocity after block 2 has collided once with block and once with the wall. Assume that the wall has infinite mass (it does not move). First look at the collision between the two blocks. Because the collision is elastic, the final velocities of the two blocks are given by the relations

8 v f = m m 2 m +m 2 v i + 2m 2 m +m 2 v 2f = 2m m++m 2 v i + m 2 m m +m 2 which simplify because v i = 0: v f = 2m 2 m +m 2 v 2f = m 2 m m +m 2 The velocity of the second block after its collision with the wall will be equal in magnitude to v 2f, but in the opposite direction; i.e., it will be equal to v 2f. Setting v f and v 2f equal, we find 2m 2 m + m 2 = m 2 m m + m 2 2m 2 = m m 2 m 2 = 3 m

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