L07_hw - Homework Chapter 07 P1 What is the magnitude of...

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Unformatted text preview: Homework Chapter 07 P1. What is the magnitude of the momentum of a 28-g sparrow flying with a speed of 8.4 m/s? Solution: kg m/s kg m/s P3. A 0.145-kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0 m/s. If the contact time between bat and ball is 3.00 s, calculate the average force between the ball and bat during contact. Solution: Choose the direction from the ball to the pitcher to be positive. Thus, the velocity after the hit will be negative, since the ball will be moving in opposite direction. N " " means the force points in opposite direction to our choice - the force points from the bat to the ball. P5. Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1500 kg/s with a speed of 4.0 m/s (at the moment of takeoff). Solution: Force on the gas: m/s Force is pointing downward. Newton's Third Law: The force on the rocket is equal in magnitude but opposite in direction to that of the gas: N pointing upwards P7. A 12,600-kg railroad car travels alone on a level frictionless track with a constant speed of 18.0 m/s. A 5350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed? Solution: Momentum along the horizontal direction is conserved. kg s kg m/s s m/s N kg m/s 1 kg kg m/s kg m/s P8. A 9300-kg boxcar traveling at 15.0 m/s strikes a second boxcar at rest. The two stick together and move off with a speed of 6.0 m/s. What is the mass of the second car? Solution: Momentum along the direction of the cars is conserved. kg m/s kg m/s m/s m/s kg P15. A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5 s. find (a) the impulse imparted to the golf ball, and (b) the average force exerted on the ball by the golf club. Solution: ! ! kg m/s s kg N m/s m/s kg m/s P17. A tennis ball of mass m = 0.060 kg and speed of v = 25 m/s strikes a wall at a 45 angle and rebounds with the same speed at 45. What is the impulse (magnitude and direction) given to the ball? Solution: y 45 x 45 Choose a coordinate system: x - perpendicular to the wall, inward direction y - along the wall, upward 2 cos cos Note: the angles are measured with respect to x! cos cos kg m/s " " " " " Change of momentum along x: kg m/s, pointing away from the wall. P19. A 95-kg fullback is running at 4.0 m/s to the east and is stopped in 0.75 s by a headon tackle by a tackler running due west. Calculate (a) the original momentum of the fullback, (b) the impulse exerted on the fullback, (c the impulse exerted on the tackler, and (d) the average force exerted on the tackler. Solution: Original direction of the fullback is east - choose as positive direction. #$ % #$ % kg #$ % m/s kg m/s kg m/s kg m/s ! kg m/s & Momentum is conserved for the system of the two persons. ' #$' &# ( # ) &* #$ % '$ &# ( # &* '$ #$ % kg m/s kg m/s kg m/s kg m/s s ) kg m/s &* &* '$ &* '$ + &* '$ &* '$ N P22. A ball of mass 0.440 kg moving east (+ x direction) with a speed of 3.30 m/s collides head-on with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision? Solution: Choose positive direction in the direction of the initial velocity of the 0.440-kg ball. velocity of the 0.440-kg ball before the collision velocity of the 0.220-kg ball before the collision velocity of the 0.440-kg ball after the collision velocity of the 0.220-kg ball after the collision Momentum is conserved: , Energy is conserved (elastic collision): 3 ,, Rewrite eqs. (*) and (**) in a more convenient form: Combining the two equations gives us: ,,, Thus, the equations that need to be solved are , and ,,, kg m/s kg kg kg m/s m/s m/s m/s m/s m/s m/s Note: the velocities turned out to have positive magnitude - they are in the same direction as the initial direction of the 0.440-kg ball. P39. A 15.0-kg object moving in the +x direction at 5.5 m/s collides head-on with a 10.0 kg object moving in the -x direction at 4.0 m/s . Find the final velocity of each mass if (a) the objects stick together; (b) the collision is elastic; (c) the 15.0-kg object is at rest after the collision; (d) the 10.0-kg object is at rest after the collision; (e) the 15.0 kg object has 4 a velocity of 4.0 m/s in the -x direction after the collision. Are the results in (c), (d) and (e) reasonable? Solution: (a) When balls stick together, the collision is not elastic. Energy is not conserved. m/s (b) Elastic. collision - the problem is identical to P22. We will use eqs. (*) and (***) Using equation (***) from P22.: . Substituting in (*) . kg m/s kg kg kg m/s m/s m/s m/s (c) m/s We don't know whether the collision is elastic. or non-elastic. . we cannot use the conservation of energy equation. The momentum is conserved, however, so: kg m/s kg kg kg kg m/s m/s Note: We might not know whether energy is conserved, but we certainly know that we cannot have more mechanical energy after the collision than what we have before. Thus, our check for "reasonableness" is to see how the energy changes: / 0 J is The energy after the collision is less than the energy before, thus a situation in which zero is possible and reasonable. (d) m/s Similarly to (c): 5 kg m/s kg kg kg kg m/s m/s Since ball A comes from behind, it is physically impossible to hit ball B, so that ball B remains still, but A continues moving. The situation is unreasonable. (e) m/s kg m/s kg kg kg kg m/s m/s The directions are reasonable. Ball A hits B and bounces back, while B continues forward. We must check how the energy changes, though! / 0 1 22 2 It appears that the balls have gained energy which is impossible. Thus, the situation is impossible, because energy is either conserved, or lost in this case. In order to gain energy, there must be a force to do work. P46. Find the center of mass of the three-mass system shown in the Fig. Specify relative to the left hand 1.00 kg mass. Solution: y 1. 00 kg 1. 50 kg 1.10 kg x 0.50 m 0.25 m Since we must give our answer with respect to the ball on the left-hand side, we put the coordinate system at the center of that ball and measure all the distances from there. "& " " " kg m kg kg m kg kg kg m m m P77. An atomic nucleus at rest decays radioactively into an -particle and a smaller nucleus. What will be the speed of this recoiling nucleus if the speed of the alpha particle is 3.8 m/s? Assume the recoiling nucleus has a mass 57 times greater than that of the alpha particle. 6 Solution: All the forces that act on the system are internal, hence, they will appear as pairs into the system force diagram. Their sum will be zero (Newton's third law) - momentum will be conserved. ' #$' +'& 3 4 4 '$ +'& 3 5 5 4 5 initial nucleus recoiling nucleus The total mass of the system is also conserved - we cannot loose it, or gain. Thus: 4 4 4 5 5 m/s m/s 5 m/s opposite to the direction of the alpha particle. 7 ...
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This document was uploaded on 09/21/2009.

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