Physics 16 Problem Set 8 Solutions

Physics  16  Problem  Set  8  Solutions   Y&F  Problems     8.20. IDENTIFY: In part (a) no horizontal force implies is constant. In part (b) use the energy expression, Eq. 7.14, to find the potential energy intially in the spring. SET UP: Initially both blocks are at rest. Figure 8.20 EXECUTE: (a) Block A has a final speed of 3.60 m/s, and moves off in the opposite direction to B. (b) Use energy conservation: . Only the spring force does work so (the blocks initially are at rest) (no potential energy is left in the spring) the potential energy stored in the compressed spring. Thus EVALUATE: The blocks have equal and opposite momenta as they move apart, since the total momentum is zero. The kinetic energy of each block is positive and doesn’t depend on the direction of the block’s velocity, just on its magnitude.   8.27. IDENTIFY: The horizontal component of the momentum of the system of the rain and freight car is conserved. SET UP: Let +x be the direction the car is moving initially. Before it lands in the car the rain has no momentum along the x axis. EXECUTE: (a) says and . (b) After it lands in the car the water must gain horizontal momentum, so the car loses horizontal momentum. EVALUATE: The vertical component of the momentum is not conserved, because of the vertical external force exerted by the track. 8.38. IDENTIFY: Apply conservation of momentum to the collision. Apply conservation of energy to the motion of the block after the collision.   SET UP: Conservation of momentum applied to the collision between the bullet and the block: Let object A be the bullet and object B be the block. Let be the speed of the bullet before the collision and let V be the speed of the block with the bullet inside just after the collision. Figure 8.38a is constant gives . V Conservation of energy applied to the motion of the block after the collision: y #1 #2 v50 A1B x 0.230 m Figure 8.38b EXECUTE: Work is done by friction so (no work done by gravity) (block has come to rest) Thus Use this in the conservation of momentum equation EVALUATE: When we apply conservation of momentum to the collision we are ignoring the impulse of the friction force exerted by the surface during the collision. This is reasonable since this force is much smaller than the forces the bullet and block exert on each other during the collision. This force does work as the block moves after the collision, and takes away all the kinetic energy.   8.41. IDENTIFY: When the spring is compressed the maximum amount the two blocks aren’t moving relative to each other and have the same velocity relative to the surface. Apply conservation of momentum to find V and conservation of energy to find the energy stored in the spring. Since the collision is elastic, Eqs. 8.24 and 8.25 give the final velocity of each block after the collision. SET UP: Let +x be the direction of the initial motion of A. EXECUTE: (a) Momentum conservation gives and . Both blocks are moving at 0.333 m/s, in the direction of the initial motion of block A. Conservation of energy says the initial kinetic energy of A equals the total kinetic energy at maximum compression plus the potential energy stored in the bumpers: (b) and . . Block A is moving in the direction at 1.33 m/s. . Block B is moving in the +x direction at 0.667 m/s. EVALUATE: When the spring is compressed the maximum amount the system must still be moving in order to conserve momentum.   8.68. IDENTIFY: The total momentum of the system is conserved and is equal to zero, since the pucks are released from rest. SET UP: Each puck has the same mass m. Let +x be east and +y be north. Let object A be the puck that moves west. All three pucks have the same speed v. EXECUTE: gives and . gives and . Since and the y components are equal in magnitude, the x components must also be equal: The angle and says . If is positive then that puck B makes with the x axis is given by and in a direction north of east and the other puck moves in a direction EVALUATE: Each component of momentum is separately conserved. is negative. . One puck moves south of east.   8.69. IDENTIFY: The x and y components of the momentum of the system are conserved. Set Up: After the collision the combined object with mass moves with velocity Solve for and EXECUTE: (a) . . gives . . . gives . . (b) . . . EVALUATE: Since there is no horizontal external force the vector momentum of the system is conserved. The forces the spheres exert on each other do negative work during the collision and this reduces the kinetic energy of the system.   8.71.   IDENTIFY: The horizontal component of the momentum of the sand plus railroad system is conserved. SET UP: As the sand leaks out it retains its horizontal velocity of 15.0 m/s. EXECUTE: The horizontal component of the momentum of the sand doesn’t change when it leaks out so the speed of the railroad car doesn’t change; it remains 15.0 m/s. In Exercise 8.27 the rain is falling vertically and initially has no horizontal component of momentum. Its momentum changes as it lands in the freight car. Therefore, in order to conserve the horizontal momentum of the system the freight car must slow down. EVALUATE: The horizontal momentum of the sand does change when it strikes the ground, due to the force that is external to the system of sand plus railroad car.