Question

A cannon shell moving along a straight line, explodes into two parts. Just after the explosion, one part moves with momentum $100\text{Ns}$ making an angle ${30}^{\circ }$ with the original line of motion. The minimum momentum of the other part of the shell after the explosion is

• A. $0\text{Ns}$
• B. $100\text{Ns}$
• C. $75\text{Ns}$
• D. $50\text{Ns}$

Answer 1

The correct option is D $50\text{Ns}$


Given, $p_1=100\text{Ns}$
Let the minimum momentum of the other part be $p_2$

Applying momentum conservation in y - direction,
$p_1\sin {30}^{\circ }=p_2\sin \theta$
$100\times \frac{1}{2}=p_2\sin \theta$
$\Rightarrow p_2\sin \theta =50$

Now, for $p_2$ to be minimum, $\sin \theta$ must be maximum i.e $\sin \theta =1$
$\Rightarrow (p_2{)}_{min}=50\text{Ns}$