Question

Two balls of equal mass undergo head-on collision while each was moving with a speed of $6m/s$. If the coefficient of restitution is $1/3$, the speed of each ball after impact will be

Answer 1

Step 1: Given data

Speed,

$u_1=6\frac{\text{m}}{\text{s}}$

$u_2=-6\frac{\text{m}}{\text{s}}$

Coefficient of restitution is $\frac{1}{3}$.

Step 2: To find

Speed of each ball after impact

Step 3: Calculate the speed of the ball

There is no external force acting, momentum will be conserved.

$$\begin{gathered} m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2 \\ m\times 6-m\times 6=m{v}_1+m{v}_2 \\ {v}_1=-v_2 \end{gathered}$$ …(1)

Here, $u_{1,}{u}_2$ are velocity of balls before impact and $v_1,v_2$ are velocity of balls after impact. $m_1,m_2$ are mass of balls.

$$\begin{gathered} e=\frac{v_2-v_1}{u_1-u_2} \\ \frac{1}{3}=\frac{v_2-v_1}{u_1-u_2} \end{gathered}$$ …(2)

Here $e$ is a coefficient of restitution.

After solving equations (1) and (2) get the values of speeds are

$$\begin{gathered} v_1=-2\frac{\text{m}}{\text{s}} \\ {v}_2=2\frac{\text{m}}{\text{s}} \end{gathered}$$

Hence, speed of balls are $-2\frac{\text{m}}{\text{s}}$ and $2\frac{\text{m}}{\text{s}}$.