G9-3. Solving Problems with both Momentum and Energy Conservation
The problem of the bowling ball and the student that will be solved below has some similarities to that of the ballistic pendulum. See Example 9-5 of the text for the solution of the ballistic pendulum problem. In the ballistic pendulum, a bullet is fired horizontally into a massive target. For the bowling ball and the student, the situation is reversed. The bowling ball swings down to be caught by the student, who stands on a cart that is free to roll.
Let's consider the ballistic pendulum first. The bullet embeds in the target, which then swings upward like a simple pendulum. The solution of the problem requires both conservation of energy and momentum. Conservation of momentum is applied to the collision. With Fnet,ext = 0 for the system of the bullet and the target, one can say the total momentum of the system is conserved: Pi = Pf.. Conservation of energy is then applied to the upward swing, which begins immediately after the bullet embeds and the target has its maximum horizontal velocity. The final state is when the target reaches its highest position. The net work done by the external force of tension on the system of the bullet and target is 0; hence, 0 = ΔK + ΔUg can be applied to the upward swing.
A common mistake in solving the problem of the ballistic pendulum is to treat the entire situation as a conservation of energy problem in which the initial state is the bullet just before colliding with the target and the final state is the highest point of the upward swing. The reason that this method is incorrect is that kinetic energy is lost in the collision, which is inelastic. The bullet experiences kinetic friction as it embeds in the target; hence, there is an external force of friction acting and Wext = Wf . In principle, one could solve the problem this way if the work done by friction could be calculated. However, there is no simple way to do that in this situation. Hence, one avoids calculating the Wf term by splitting the problem into two parts. In applying conservation of momentum to the collision, the frictional force is not external but is rather the very force that results in momentum changes in the collision.
The two example problems below are solved using the two-part method.
The following calculation shows that kinetic energy is lost as the student catches the bowling ball.
The initial kinetic energy (just before the ball is caught) is all in the ball.
The final kinetic energy (just after the ball is caught) is shared by the ball, student, and cart.
The fractional change of kinetic energy is calculated below.
If kinetic energy were conserved, the result would be 0. In actuality, the result is always negative, indicating a loss of kinetic energy. The fractional loss depends on the ratio of the mass of the student + cart to the mass of the system.
In moving from X to Y on a level surface, Block A encounters no friction as it moves toward stationary Block B at constant velocity, vo, as shown to the right. The two blocks of equal mass M stick together in the collision. During the small time t that the collision lasts, the blocks move a negligible distance. After the collision, the blocks slide together through and beyond region YZ. In region YZ, which is a distance d in length, the track exerts a constant friction force f to the left on the combined blocks. Express answers in simplest form in terms of the given symbols M, vo, d, f, and t.
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