Problem on inelastic collisions and coefficient of restitution, conservation of

Question

Problem on inelastic collisions and coefficient of restitution, conservation of energy?

When two pool balls collide head-on, the ratio of their relative speeds after the collision to their relative speeds before the collision is |v2 - v1|a / |v2 - v1|b R, called the coefficient of restitution. (R = 1 if the collision is elastic, as explained in Section 13.1.) (a) In such a collision of equal-mass balls in which one ball is initially at rest and the other has velocity V0, find the final velocities v1 (of the incident ball) and v2 (of the target ball) in terms of v0 and R. (b) Find the fractional change in the total kinetic energy, (K0-Kf) / K0, in terms of R, where Kf is the sum of the final kinetic energies. (c) Check your results in the limiting cases R = 0 and R = 1.

Explanation / Answer

a) by balancing momentum we have,

mv0+m*0=mv1 + mv2

v0=v1+v2 (1)

R=v2-v1/v0-0

or, Rv0=v2-v1 (2)

Adding equatiom (1) & (2)

2v2=v0(1+R)

v2=v0(1+R)/2

& v1= v0(1-R)/2

b) initial kinetic energy(K0) = 0.5mv0^2

final kinetic energy(Kf)=0.5m v1^2 + 0.5mv2^2

=0.5m(v1^2 + v2^2)

=0.5mv0^2*((1-R)/2)^2 + ((1+R)/2))^2)

(K0-Kf)/K0

=1 - Kf/K0

=1- [((1-R)/2)^2 + ((1+R)/2))^2)]

c)for, R=0

(K0-Kf)/K0 = 1/2.

for R=1,

(K0-Kf)/K0 = 0.

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