7. We found that if a cue ball, rolling without slipping at speed v0, strikes an

ID: 1390822 • Letter: 7

Question

7. We found that if a cue ball, rolling without slipping

at speed v0, strikes an identical, stationary billiard ball head-on, eventually both balls will

roll without slipping. The balls are uniform solid spheres, each of mass m, radius r, and

moment of inertia I =2/5 m r^2 about its center. The final speed of the target ball is 5/7 *v0 ; that

of the cue ball is 2/7 v0. Calculate the total fraction of the initial kenetic energy of the cue

ball dissipated by frictional forces between the felt of the table and the two balls spinning or

sliding, before they finally attain rolling without slipping at their respective final speeds.

The work done by the cue ball is equal to change in kinetic energy. So we can write

W = dKE = KE2 - KE1

= [0.5 mv22 + 0.5 I2 w'22 ] + [ 0.5mv12 + 0.5 I1w'12 ] - { [0.5 mu22 + 0.5 I2 w22 ] + [ 0.5mu12 + 0.5 I1w12 ] }

Given that I1 = I2 = I = (2/5 ) m r2

u1 = vo , u2 = 0

v1 = (2/7) vo , v2 = (5 / 7) vo

w1 = u1 / r , w2 = 0

w1' = (2/7) (vo / r), w2' = (5/7) (vo / r)

KE1i = [ 0.5mv12 + 0.5 Iw'12 ] = [0.5 m vo2 + 0.5 * (2/5) * m r2 * (vo / r)]2 ] = 0.7 mvo2

KE2i = 0

KE1f = [0.5 m[(2/7) (vo)]2 + 0.5 * (2/5) * m r2 * [(2/7) (vo / r)]2 ] = 0.0816 mvo2 + 0.0163 * mvo2 = 0.0979mvo2

KE2f = [0.5 m[(5/7) (vo)]2 + 0.5 * (2/5) * m r2 * [(5/7) (vo / r)]2 ] = 0.255mvo2 + 0.102 mvo2 = 0.357mvo2

KE1f + KE2f = 0.4549 mvo2

Then the total fraction of the initial kenetic energy of the cue ball dissipated by frictional forces is

[KE1f + KE2f ] / KE1i = 0.4549 mvo2 / 0.7 mvo2 = 0.649 or 64.9%

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