Question

A ball of mass m moving with a speed $2v_{0}$collides head-on with an identical ball at rest. If e is the coefficient of restitution, then what will be the ratio of velocity of two balls after collision?

• A. $\frac{1 - e}{1 + e}$
• B. $\frac{1 + e}{1 - e}$
• C. $\frac{e - 1}{e + 1}$
• D. $\frac{e + 1}{e - 1}$

Answer 1

Answer: (A)

$\frac{1 - e}{1 + e}$

Answer 2

According to the law of conservation of linear momentum, we get

$m(2v_{0}) + m \times 0 = mv_{1} + mv_{2}$

Where $v_{1}andv_{2}$are the velocities of the balls after collision.

$v_{1} + v_{2} = 2v_{0}$ ….(i)

By definition of coefficient of restitution

$e = \frac{v_{2} - v_{1}}{u_{1} - u_{1}} = \frac{v_{2} - v_{1}}{2v_{0}}$ $(\therefore u_{1} = 2v_{0,}u_{2} = 0)$

$v_{2} - v_{1} = 2ev_{0}$ ……(ii)

Adding (i) and (ii), we get

$2v_{2} = 2v_{0} + 2ev_{0}$

$v_{2} = (1 + e)v_{0}$

Subtract (ii) from (i), we get

$2v_{1} = 2v_{0} - 2ev_{0}$

$v_{1} = v_{0}(1 - e)$

Their corresponding ratio is

$\frac{v_{1}}{v_{2}} = \frac{1 - e}{1 + e}$