A shell is fired from a cannon with velocity v m/s at an angle θ with the horizontal direction. At the highest point in 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/s of the other piece immediately after the explosion is:

Question

A shell is fired from a cannon with velocity v m/s at an angle θ with the horizontal direction. At the highest point in 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/s of the other piece immediately after the explosion is: (A) 3v cos θ (B) 2v cos θ (C) (3v/2) cos θ (D) (√3v cos θ/2)

Tardigrade Admin · Accepted Answer

In case of projectile motion as at the highest point

(v)_{vertical} = 0

and

(v)_{horizontal} = v cos θ

the initial linear momentum of the system will be

m v cos θ

Now as force of blasting is internal and force of gravity is vertical, so linear momentum of the system along horizontal is conserved, i.e.,

P_{1} + P_{2} = m v cos θ

or

m_{1} v_{1} + m_{2} v_{2} = m v cos θ

But it is given that

m_{1} = m_{2} = m /2

and as one part retraces its path,

V_{1} = - V cos θ

\frac{1}{2} m (- v cos θ) + \frac{1}{2} m v_{2}

= m v cos θ

Solving, we get:

v_{2} = 3 v cos θ

$3 v cos θ$

$2 v cos θ$

$\frac{3 v}{2} cos θ$

$\frac{3 v cos θ}{2}$

3vcosθ

2vcosθ

23v​cosθ

23​vcosθ​

$3 v cos θ$

$2 v cos θ$

$\frac{3 v}{2} cos θ$

$\frac{3 v cos θ}{2}$