A body x with a momentum p collides with another identical stationary body y one dimensionally. During the collision y gives an impulse J to the body x. Then, the coefficient of restitution is:

Question

A body x with a momentum p collides with another identical stationary body y one dimensionally. During the collision y gives an impulse J to the body x. Then, the coefficient of restitution is: (A)  (2j/p)-1  (B)  (j/p)+1  (C)  (j/p)-1  (D)  (j/2p)-1

Tardigrade Admin · Accepted Answer

Coefficient of restitution e=(v2-v1/u1-u2) Here both the bodies are identical i.e., have the same mass, So, e=(mv2-mv1/mv1-μ 2) =(P2-P1/p1-p2) p1=p (Intial momentum of first body p2= Initial momentum of second body) = 0 ( ∵ Final momentum p2=p+J ) ( ∴ Impulse = change in momentum) p1=0 ( ∵ When two bodies of equal masses collide elastically then thy exchange then velocities) ∴ e=(p+J/p) =1+(J/p)

Q. A body x with a momentum p collides with another identical stationary body y one dimensionally. During the collision y gives an impulse J to the body x. Then, the coefficient of restitution is:

$\frac{2 j}{p} - 1$

$\frac{j}{p} + 1$

$\frac{j}{p} - 1$

$\frac{j}{2 p} - 1$

Q. A body x with a momentum p collides with another identical stationary body y one dimensionally. During the collision y gives an impulse J to the body x. Then, the coefficient of restitution is:

p2j​−1

pj​+1

pj​−1

2pj​−1

$\frac{2 j}{p} - 1$

$\frac{j}{p} + 1$

$\frac{j}{p} - 1$

$\frac{j}{2 p} - 1$