Question

a ball with mass m1=1kg is released from a height of 50m while another ball with mass m2=1.0kg is launched vertically towards the first ball with v=25m/s consider that their collision is perfectly elastic. What is the velocity of the first ball immediately after collision (neglect friction) A. -5.4m/s B. 5.4m/s C. -19.6 m/s D. 19.6 m/s

Answer 1

Let y be the height from the ground where the collision occurs.

then, for m1:

[50 - h] = ut + (1/2)gt² = (1/2)gt²

=> t² = ([100 - 2h]/9.8) = 10.204 - 0.2041h

in the same time, the vertical distance covered by m2 is:

h = 25t - (1/2)gt² = 25t - 4.9t²

substitute this above

t² = 10.204 - 5.102t + t²

=> t = 2 s

for t = 2 s, velocity of the m2 ball is:

v = 25 - 9.8(2) = 5.4 m/s

now, since the collision is perfectly elastic and here m₁ = m₂, the velocity of m2 before the collision = velocity of m1 after the collision.

Therefore, velocity of the first ball m1 immediately after the collision is: v' = +5.4 m/s.