Question

A moving body of mass $m_1$ strikes a stationary body of mass $m_2$. The masses $m_1$ and $m_2$ should be in the ratio $\frac{m_1}{m_2}$ so as to decrease the velocity of the first body 1.5 times assuming a perfectly elastic impact. Then the ratio $\frac{m_1}{m_2}$ is

• A. $\frac{1}{25}$
• B. $\frac{1}{5}$
• C. $5$
• D. $25$

Answer 1

The correct option is C $5$

Let initial velocity of mass $m_1$ be $u_1$ and its final velocity be $v_1$
the initial velocity for mass $m_2$ is zero and let it's final velocity be $v_2$,

According to law of conservation of momentum ,
$m_1\times u_1+m_2\times u_2=m_1\times v_1+m_2\times v_2$
or
$m_1\times u_1=m_1\times v_1+m_2\times v_2$ (since, $u_2$ $=0$)

According to the equation for $\text{coefficient of restitution for a perfectly elastic collision}$, we have
$v_2-v_1=\left(u_1\right)$
solving both the equations gives the relation
$\frac{(m_1-m_2)}{(m_1+m_2)}=\frac{v_1}{u_1}=\frac{2}{3}$

Applying componendo and dividendo, we have
$\frac{m_1}{m_2}=\frac{3+2}{3-2}=\frac{5}{1}$
$\frac{m_1}{m_2}=5$