Question

A shell is fired from a cannon with velocity v m/sec at an angle 0 with the horizontal direction' At the highest point it its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed in m/sec of the other piece immediately after the explosion is

Moderate

Solution

As the shell is at the highest point, hence the velocity of the shell before explosion must be entirely horizontal i.e. v cos $θ .$ Now when the explosion takes place, the total momentum before and after the collision must be the same( explosion consists of only internal forces). The piece which retraces its path to the cannon has velocity -v cos $θ .$ immediately after collision.Let P₂ be the momentum of second particle, then
$p_{2} - \frac{m}{2} vcos θ = mvcos θ ∴ p_{2} = mvcos θ + \frac{m}{2} vcos θ If v_{2} be the velocity of the second piece, then \frac{m}{2} \times V_{2} = (3 / 2) mvcos θ v_{2} = 3 vcos θ$

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