Question

A projectile is fired with velocity u at angle $\theta$ with horizontal. At the highest point of its trajectory it splits up into three segment of masses $m,m$ and $2m$. First part falls with zero initial velocity vertically downward and second part return back via same path to the point of projection. Then just after explosion, velocity of third part of mass 2m will be

• A. $u\cos \theta$
• B. $\frac{3}{2}u\cos \theta$
• C. $2u\cos \theta$
• D. $\frac{5}{2}u\cos \theta$

Answer 1

The correct option is D $\frac{5}{2}u\cos \theta$

Given, $velocity=v,Angle=\theta ,masses=m,m,2m$

So, As the highest point, the velocity of the projectile is $u\cos \theta$

Now, the projectile has mass $4m$ so the momentum possessed by the projectile at the highest point $=4mu\cos \theta$ (Right Direction)

It breaks into three fragments, one of the fragment falls down so its horizontal momentum is zero.

The second fragment of the mass m return to the same path so it must have the velocity of $u\cos \theta$ in the opposite direction (left direction)

So,

The fragments of the third fragment of mass 2m, Let the velocity of the third fragment is v,

Thus the momentum of the third fragment is $2mv$

Conserving momentum in horizontal dirction,

We have,

$p_i=p_f\Rightarrow 4mu\cos \theta =-mu\cos \theta +2mv\Rightarrow V=\frac{5}{2}u\cos \theta$