Question

A ball of mass m moving at a speed v makes a head-on collision with an identical ball at rest. The kinetic energy of the ball after the collision is three-fourths of the original. Find the coefficient of restitution.

Answer 1

Step 1: Given data

Mass of 1st ball = m

Speed = v

Mass of 2nd ball = m

Let the final velocities of 1st and 2nd ball are v₁ and v₂ respectively.

Step 2: Formula used

Law of conservation of momentum

$m(v_1+v_2)=mv$

Step 3: Calculating the coefficient of restitution

Using the law of conservation of momentum,

$m(v_1+v_2)=mv$ ...(i)

v₁+v₂=v ...(ii)

also v₁−v₂=ev

Given that final K.E.=$\frac{3}{4}$ initial K.E.

$\frac{1}{2}{m}_1{v_1}^2+\frac{1}{2}{m}_2{v_2}^2=\frac{3}{4}\left(\frac{1}{2}m{v}^2\right)$

${v_1}^2+{v_2}^2=\frac{3}{4}{\left(v\right)}^2$

$\frac{{\left(v_1+v_2\right)}^2+{\left(v_1-v_2\right)}^2}{2}=\frac{3}{4}{\left(v\right)}^2$

$\frac{\left(1+e^2\right)v^2}{2}=\frac{3}{4}{\left(v\right)}^2$

$\left(1+e^2\right)=\frac{3}{2}$

$e^2=\frac{1}{2}$

$e=\frac{1}{\sqrt{2}}$

Hence, the value of the coefficient of restitution is $\frac{1}{\sqrt{2}}$