Question

Two metal balls of masses $10\text{kg}$ and $20\text{kg}$, moving in opposite directions on a frictionless floor, collide elastically. If the velocities of the balls are $3\text{m/s}$ and $1\text{m/s}$ respectively before collision as shown in the figure, then the velocities of the balls after collision are respectively

• A. $\frac{5}{3}\text{m/s}$ towards left, $\frac{7}{3}\text{m/s}$ towards left
• B. $\frac{5}{3}\text{m/s}$ towards right, $\frac{7}{3}\text{m/s}$ towards right
• C. $\frac{7}{3}\text{m/s}$ towards left, $\frac{5}{3}\text{m/s}$ towards right
• D. $\frac{5}{3}\text{m/s}$ towards left, $\frac{7}{3}\text{m/s}$ towards right

Answer 1

The correct option is C $\frac{7}{3}\text{m/s}$ towards left, $\frac{5}{3}\text{m/s}$ towards right

Given,
Mass of metal ball 1 $\left(m_1\right)=10\text{kg}$
Mass of metal ball 2 $\left(m_2\right)=20\text{kg}$
Speed of metal ball 1 before collision $\left(u_1\right)=3\text{m/s}$
Speed of metal ball 2 before collision $\left(u_2\right)=1\text{m/s}$

Let the velocities after collision be $v_1$ and $v_2$.
Using law of conservation of momentum (taking rightward +ve):
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$10\left(3\right)+20(-1)=10v_1+20v_2$
$\Rightarrow v_1+2v_2=1...\left(1\right)$
For elastic collision, $e=1$
$\Rightarrow v_2-v_1=u_1-u_2$
$\Rightarrow v_2-v_1=4...\left(2\right)$

From (1) and (2), we get
$3v_2=5\Rightarrow v_2=\frac{5}{3}\text{m/s}$ (right)
& $v_1=1-2v_2=1-\frac{10}{3}=\frac{-7}{3}\text{m/s}$ (left)