Question

A cannon shell moving along a straight-line burst into two parts. Just after the burst one part moves with momentum $40Ns$ making an angle $30^o$ with the original line of motion. What will be the minimum momentum of the other part of shell just after the burst ?

• A. $0Ns$
• B. $10Ns$
• C. $20Ns$
• D. $17.32Ns$

Answer 1

Answer: (C)

As shown in the figure, the components of momentum of one shell along initial direction and perpendicular to initial direction are $P_{1x}=20\sqrt{3}Ns$ and $P_{1x}=20\sqrt{3}Ns$

For momentum of the system to be zero in y-direction, $P_{2y}$ must be $20Ns$ . $2^{nd}$ part of shell may or may not have momentum in $x$ -direction
$\therefore P_{2min}=20N/s$