Submitting assignments Schedules Course Info Ch. Reviews Guides Problems Labs Videos AP Info Announcements Contact Shortcuts WebAssign Canvas Equations Lab FAQ Software IWP

G9-3. Solving Problems with both Momentum and Energy Conservation

Introduction

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.

Example 1

 Problem statement, given, and strategy A student stands on a low-friction wheeled cart as shown to the right. A bowling ball suspended by a string from the ceiling is pulled to an angle of 60° away from the vertical and released. The ball swings down and is caught by the student. What is the velocity of the student, cart, and ball immediately after the collision? Given:  mb = mass of bowling ball ms = mass of student + cart vib = initial velocity of bowling ball = 0 L = length of string θ = 60° Strategy: Part A. Use conservation of energy to solve for the velocity of the ball just before being caught. Part B. Use conservation of momentum to solve for the velocity of the student with the ball just after the collision ends. Part A. Using conservation of energy for the downward swing Make these additional definitions: vfb = velocity of the bowling ball at the bottom of the swing but just before contact with the student yi = the initial height of the bowling ball = L - Lcosθ. yf = the final height of the bowling ball = 0 Take the system to be the ball and the Earth. Tension of the string is an external force which does no work on the ball. initial state:  ball being released final state: ball just before it's caught energy changes: ΔK > 0; ΔUg < 0 Part B. Using conservation of momentum for the collision Make these additional definitions: vis = initial velocity of the student and cart = 0 vf' = final velocity of the student, cart, and ball immediately after the collision Define the system to be the ball, student, and cart. Tension of the string on the ball, weight of the ball, student, and cart, and normal force on the cart are external forces which all add to 0; therefore, momentum is conserved. initial state:  ball just before it's caught final state: ball just after it's caught Substitute the expression for vfb from Part A into the result for Part B to obtain the desired result in terms of the given quantities.

Reflection

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.

Example 2

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.

1. Determine the velocity v of the combined blocks immediately after the collision. (This is a conservation of momentum problem.)

System: blocks A and B

External forces: Normal and gravity add to 0 so Fnet,ext = 0, and momentum is conserved. We’re assuming that in the short time that the collision lasts, friction doesn’t have a significant effect.

States: initial – just before collision; final – just after collision

1. What percentage of block A’s kinetic energy before the collision do the combined blocks have immediately after the collision?

We’re looking for the quantity 100Kf/Ki.

This result indicates that total kinetic energy is lost in the collision; therefore, the collision is inelastic. Like the problems of the ballistic pendulum and the bowling ball and the student, the loss of kinetic energy in the collision is why we must do this problem using the two part conservation of momentum and energy method.

1. Determine the velocity vz of the combined blocks at Point Z. (This can be done using strictly net force-dvat methods, but we’ll use energy methods, since that’s our topic.)

System: blocks A and B

External forces: Normal and gravity do no work since the forces are perpendicular to the displacement of the blocks. However, kinetic friction does work on the blocks.

States: initial – blocks just after the collision; final - blocks reaching point Z

Energy changes: K decreases. There is no potential energy.

The work done by friction is Wf = fdcosθ, where θ = 180°. Therefore, Wf = -fd.

A check on this result is to examine the case of no friction, f = 0. In that case, vz = vo/2. This makes sense, because the blocks wouldn’t slow down in the absence of friction.