Figure 8.50
Problem 65.
8.65. Set Up: Apply conservation of momentum to the collision and conservation of energy to the motion after the collision. Let +x be to the right.
Solve: Collision: Pi,x = Pf,x gives (4.20 × 10−3kg)v = (4.20 × 10−3kg + 2.50 kg)v′
Motion after the collision: The kinetic energy of the block immediately after the collision is converted entirely to gravitational potential energy at the maximum angle of swing. The figure below shows that the maximum height h the block swings up is given by h = (0.750 m)(l − cos34.7°) = 0.133 m.
Conservation of energy gives
Then the conservation of momentum equation gives
Reflect: The original speed of the bullet is nearly three times the speed of sound in air. Kinetic energy is not conserved in the collision. Our analysis assumes that the block moves very little during the collision, while the bullet is coming to rest relative to the block. This is a good approximation since the velocity the block gets in the collision is much less than the initial velocity of the bullet.
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