Question

A cannon shell moving along a straight line bursts into two parts. Just after the burst one part moves with momentum 20 N s making an angle ${30}^o$ with the original line of motion. The minimum momentum of the other part of shell just after the burst is :

• A. 0 N s
• B. 5 N s
• C. 10 N s
• D. 17.32 N s

Answer 1

The correct option is C 10 N s

Given : $P_1=20$ Ns

Let the minimum momentum of other part be $P_2$

Applying conservation of momentum in y direction :

$P_{1}sin30=P_{2}sin\theta$

$20\times \frac{1}{2}=P_{2}sin\theta$

$⟹P_{2}sin\theta =10$ N s

Now for $P_2$ to be minimum, $sin\theta$ must be maximum i.e $sin\theta =1$

$⟹P_2=10$ N s