Question

An object of mass $5\, kg$ is projected with a velocity of $20 \,ms^{-1}$ 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 $1\, kg$ and $4\, kg$. 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. $11\,m$
• B. $22\,m$
• C. $44\,m$
• D. $66\,m$

Answer 1

Answer: (C)

Here, $m = 5\, kg$, $v = 20 \,ms^{-1}$, $\theta = 60^{\circ}$
Horizontal component of velocity,
$v_{x} = v\,sin\,60^{\circ} = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\,m/s$
Time taken to reach the highest point $=$ time taken to reach the ground from the highest point
$t = \frac{v\,sin\,\theta}{g} = \frac{v_{y}}{g} = \frac{10\sqrt{3}}{9.8} = 1.77\,s$
At the highest point, m splits up into two parts masses $m_{1} = 1 \,kg$ and $m_{2} = 4\, kg$. If their velocities are $v_{1}$ and $v_{2}$ respectively, then applying the principle of conservation of linear momentum, we get
$m_{1}v_{1} + m_{2}v_{2} = mv\, cos \,\theta \quad\ldots\left(i\right)$
$v_{1}+4v_{2} = 5 \times 10 = 50$
Initial $K.E. =\frac{1}{2}\,m\,\left(v\,cos\,\theta\right)^{2} = \frac{1}{2} \times 5\left(10\right)^{2} = 250\,J
Final K.E. = 2 \left(initial K.E.\right) = 2 \times 250 = 500\, J$
$\therefore \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v^{2}_{2} = 500$
$\frac{1}{2}\times1v_{1}^{2} + \frac{1}{2}4v_{2}^{2} = 500$
or $v^{2}_{1} + 4v^{2}_{2} = 1000\quad\ldots\left(ii\right)$
On solving $\left(i\right)$ and $\left(ii\right)$, we get $v_{1} - 30\, m/s, v_{2} = 5 \,m/s$
$\therefore $ Separation between the two fragments
$= \left(v_{1} - v_{2}\right)\times t = \left(30 - 5\right) \times 1.77 = 44.25 \,m$.