Question

A bomb of mass $m$ is thrown with velocity $v$. It explodes and breaks into two fragments. One fragment of mass $m^{\prime }$, immediately after the explosion, comes to rest. The speed of the other fragment will be :

• A. $\frac{mv}{m-m^{\prime }}$
• B. $\frac{mv}{m^{\prime }}$
• C. $v$
• D. $\frac{mv}{m+m^{\prime }}$

Answer 1

The correct option is (A) $\frac{mv}{m-m^{\prime }}$

Given,

Mass of the bomb = m

The velocity of the bomb = v

After explosion

The velocity of the second fragment = $v_1$

In the given situation no external force is working so the momentum of the system will be conserved

So

$mv=(m^{\prime }\times 0)+(m-m^{\prime })\times v_1$

After solving

$v_1=\frac{mv}{(m-m^{\prime })}$

Hence the velocity of the other fragment is $\frac{mv}{m-m^{\prime }}$