Question

A ball with mass m1= 1kg is released from a height of 50 m while another ball with mass m2=1.0 kg is launched vertically toward the first ball with v0= 25 m/s. Consider that their collision is perfectly elastic what is the velocity of the first ball immediately after collision? (Neglect friction.)

Answer 1

Here,

Now, for the time taken to collide

Time taken = distance/initial relative velocity

Time taken = 50/(25) = 2 s

Now, as after the collision velocity will be exchanged

hence , the velocity of first ball just after collision = velocity of second ball just before collision

Using first equation of motion

velocity of first ball just after collision = u + a * t

velocity of first ball just after collision = 25 - 10 * 2

velocity of first ball just after collision = 5 m/s

the velocity of first ball just after collision is 5 m/s