Question

A body of mass $5 \,kg$ makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to $\frac{1}{10}$ th of its original velocity.
Then the mass of the second body is

• A. 4.09 kg
• B. 0.5 kg
• C. 5 kg
• D. 5.09 kg

Answer 1

Answer: (A)

Mass of the first body $m_1 = 5$ kg and for elastic collision coefficient of restitution, $e = 1. $

Let initially body $m_1$ moves with velocity v after collision velocity becomes $\left( \frac{u}{10}\right) $.
Let after collision velocity of M block becomes $(v_2)$ .
By conservation of momentum
$m_1u_1 + m_2 u_2 = m_1v_1 + m_2 v_2$
or $5u + M\times 0 = 5 \times \frac{u}{10} + Mv_2 $
or $5u = \frac{u}{3} + Mv_2$ ....(i)
Since , $ v_1 - v_2 = - e(u_1 - u_2) $
or $\frac{u}{10} - v_2 = - 1(u) $ or $\frac{u}{10} + u = v_2$
or $\frac{11 u}{10} =v_2$ ......(ii)
Substituting value of $v_2$ in Eq. (i) from Eq. (ii)
$5u = \frac{u}{2} +M \left(\frac{11u}{10}\right)$
or $ 5 - \frac{1}{2} = M \left(\frac{11}{10}\right)$
$\Rightarrow M = \frac{9 \times10}{2 \times11} = \frac{45}{11} $
$= 4.09 \,kg$