Question

A ball starts falling under the effect of gravitational force from a height of $45 \, m$ . When it reaches a height of $25 \, m$ it explodes into two pieces of mass ratio $1:2$ . There is no change in the vertical motion of the pieces after the explosion but they acquire horizontal velocity. If the heavier piece gains a horizontal velocity of $10 \, m \, s^{- 1}$ , then the distance between the two pieces when both of them strike the ground is

• A. $30 \, m$
• B. $10m$
• C. $20m$
• D. $15m$

Answer 1

Answer: (A)

Let us assume the mass of the ball is $3m$ and the velocity of the lighter piece in horizontal direction after the explosion is $v$ . We know that just after the collision, the horizontal velocity of the heavier piece is $10ms^{- 1}$ . Using conservation of momentum in the horizontal direction we get
$mv=2m\times 10$
$\Rightarrow v=20ms^{- 1}$
Since there is no change in vertical motion, the time taken by the pieces to reach the ground can be calculated in the following way.
Total time to fall through $45m$ is
$T_{45}=\sqrt{\frac{2 \times 45}{10}}=3s$
The time taken to fall through first $20m$ is
$T_{20}=\sqrt{\frac{2 \times 20}{10}}=2s$ .
Hence time taken by the pieces to fall from $25m$ height to ground is
$T_{40}-T_{20}=3-2=1s$
The relative velocity of the pieces in the horizontal direction is
$v_{rel}=v-\left(- 10\right)=30ms^{- 1}$
The horizontal distance between the two pieces at the time of striking the ground is
$x=30\times 1=30m$