Question

An object of mass $5kg$ is projected with a velocity of $20m{s}^{-1}$ at an angle of ${60}^{\circ }$ 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)

$44m$

Here, $m=5kg,v=20m{s}^{-1},\theta ={60}^{\circ }$
Horizontal component of velocity,
$v_x=v\cos {60}^{\circ }=20\times \frac{1}{2}=10m/s$
Vertical component of velocity,
$v_y=v\sin {60}^{\circ }=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}m/s$
Time taken to reach the highest point $=$ time taken to reach the ground from the highest point
$t=\frac{v\sin \theta }{g}=\frac{v_y}{g}=\frac{10\sqrt{3}}{9.8}=1.77s$
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
$m_1{v}_1+m_2{v}_2=mv\cos \theta ....\left(i\right)$
$v_1+4v_2=5\times 10=50$
Initial K.E. $=\frac{1}{2}m(v\cos \theta {)}^2=\frac{1}{2}\times 5(10{)}^2=250J$
Final K.E. $=2$ (initial K.E.) $=2\times 250=500J$
$\therefore \frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=500$
$\frac{1}{2}\times 1v_1^2+\frac{1}{2}4v_2^2=500$ or $v_1^2+4v_2^2=1000....\left(ii\right)$
On solving (i) and (ii), we get $v_1=30m/s,v_2=5m/s$
$\therefore$ Separation between the two fragments
$=(v_1-v_2)\times t=(30-5)\times 1.77=44.25m$.