Question

An object of mass $5kg$ is projected with a velocity of $20m{s}^{-1}$ at an angle of $60^{°}$ to the horizontal. At the highest point of its path, the projectile explodes and breaks up into two fragments of masses $1kg$ and $4kg$. The fragments separate horizontally after the explosion, which releases internal energy such that $K.E.$ of the system at the highest point is doubled. Find the separation between the two fragments when they reach the ground.

• A. $11m$
• B. $22m$
• C. $44m$
• D. $66m$

Answer 1

Answer: (C)

Here, $m=5kg$, $v=20m{s}^{-1}$, $\theta =60^{\circ }$
Horizontal component of velocity,
$v_x=vsin60^{\circ }=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}m/s$
Time taken to reach the highest point $=$ time taken to reach the ground from the highest point
$t=\frac{vsin\theta }{g}=\frac{v_y}{g}=\frac{10\sqrt{3}}{9.8}=1.77s$
At the highest point, m splits up into two parts masses $m_1=1kg$ and $m_2=4kg$. If their velocities are $v_1$ and $v_2$ respectively, then applying the principle of conservation of linear momentum, we get
$m_1{v}_1+m_2{v}_2=mvcos\theta \dots \left(i\right)$
$v_1+4v_2=5\times 10=50$
Initial $K.E.=\frac{1}{2}m{\left(vcos\theta \right)}^2=\frac{1}{2}\times 5{\left(10\right)}^2=250J<br>FinalK.E.=2\left(initialK.E.\right)=2\times 250=500J$
$\therefore \frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=500$
$\frac{1}{2}\times 1v_1^2+\frac{1}{2}4v_2^2=500$
or $v_1^2+4v_2^2=1000\dots \left(ii\right)$
On solving $\left(i\right)$ and $\left(ii\right)$, we get $v_1-30m/s,v_2=5m/s$
$\therefore$ Separation between the two fragments
$=\left(v_1-v_2\right)\times t=\left(30-5\right)\times 1.77=44.25m$.