A shell is fired from a cannon with a 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 ( m / s ) of the other piece immediately after the explosion is

Question

A shell is fired from a cannon with a 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 ( m / s ) of the other piece immediately after the explosion is (A) 3 v cos θ (B) 2 v cos θ (C)  ( 3/2) v cos θ (D)  √ ( 3/2) v cos θ

Tardigrade Admin · Accepted Answer

Let v' be the velocity of second fragment. From conservation of linear momentum, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8612c57c67f25d3c7ac6b65df1d67f9c-.png" /> 2 m(v cos θ)=m v'-m(v cos θ) ∴ v prime=3 v cos θ

$3 v cos θ$

$2 v cos θ$

$\frac{3}{2} v cos θ$

$\frac{3}{2} v cos θ$

Let $v^{'}$ be the velocity of second fragment. From conservation of linear momentum,

$2 m (v cos θ) = m v^{'} - m (v cos θ)$
$∴ v^{'} = 3 v cos θ$

3vcosθ

2vcosθ

23​vcosθ

23​​vcosθ

Let v′ be the velocity of second fragment. From conservation of linear momentum, 2m(vcosθ)=mv′−m(vcosθ) ∴v′=3vcosθ

$3 v cos θ$

$2 v cos θ$

$\frac{3}{2} v cos θ$

$\frac{3}{2} v cos θ$

Let $v^{'}$ be the velocity of second fragment. From conservation of linear momentum,

$2 m (v cos θ) = m v^{'} - m (v cos θ)$
$∴ v^{'} = 3 v cos θ$