Q. A cannon shell moving along a straight-line burst into two parts. Just after the burst one part moves with momentum $40 \, Ns$ 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 ?
NTA AbhyasNTA Abhyas 2020
Solution:
As shown in the figure, the components of momentum of one shell along initial direction and perpendicular to initial direction are $P_{1 x}=20\sqrt{3}Ns$ and $P_{1 x}=20\sqrt{3}Ns$
For momentum of the system to be zero in y-direction, $P_{2 y}$ must be $20 \, Ns$ . $2^{nd}$ part of shell may or may not have momentum in $x$ -direction
$\therefore P_{2 min}=20 \, N/s$
For momentum of the system to be zero in y-direction, $P_{2 y}$ must be $20 \, Ns$ . $2^{nd}$ part of shell may or may not have momentum in $x$ -direction
$\therefore P_{2 min}=20 \, N/s$