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 ms -1 , then the distance between the two pieces when both of them strike the ground is
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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 x 10
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 45 m is
The time taken to fall through first 20 m is
Hence time taken by the pieces to fall from 25 m height to ground is
T 40 - T 20 = 3 - 2 = 1 s
The relative velocity of the pieces in the horizontal direction is
v rel = v - (-10) = 30ms -1
The horizontal distance between the two pieces at the time of striking the ground is
x = 30 x 1 = 30 m
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