Question

A particle of mass $2m$ is projected at an angle of ${45}^{\circ }$ with horizontal with a velocity of $20\sqrt{2}{\text{ms}}^{-1}$. After $1\text{s}$ explosion takes place and the particle is broken into two equal pieces. As a result of the explosion, one part comes to rest. Find the maximum height $\left(\text{in metre}\right)$ from initial point of the particle, attained by the other part is
[Take $g=10{\text{ms}}^{-2}$]

Answer 1

Given: $v=20\sqrt{2}{\text{ms}}^{-1}$ makes an angle ${45}^{\circ }$ with the horizontal.

$v_x=v\cos {45}^{\circ }=20{\text{ms}}^{-1};v_y=v\sin {45}^{\circ }=20{\text{ms}}^{-1}$

After $1\text{s}$,

$v_{1x}=20{\text{ms}}^{-1}$ and

$v_{1y}=u_y-gt=20-(10\times 1)=10{\text{ms}}^{-1}$

$S_y=u_{y}t-\frac{1}{2}g{t}^2=20-(5\times 1)=15\text{m}$

$\therefore h=15\text{m}$

After the explosion, one pieces comes to rest, the other piece continues to move in the same direction. Since there is no external force involved in explosion. Thus, momentum will be conserved.

Now, conserving momentum in vertical direction.

$2m\times 10=(m\times 0)+m{v}_{1y}$

$\therefore v_{1y}=20{\text{ms}}^{-1}$

Height attained after the explosion is,

$h_{max}=\frac{v_{1y}^2}{2g}=\frac{(20{)}^2}{2g}=20\text{m}$

Hence, the maximum height attained by the particle will be

$H=h+h_{max}=15+20=35\text{m}$