IIT JEE Advanced 2011 Paper 2

We have
s = ut+

1

2

at2
In vertical direction, s = 5 m, u = 0 and a = g = 10 m/s. Therefore,
5m =

1

2

10t2 ⇒ t = 1s
For the ball, the horizontal displacement is 20 m. Therefore,
20m = vball × t ⇒ vball = 20 m/s
or Pball = 0.2 kg × 20 m/s = 4 kg m/s
Similarly, Pbullet = 1 kg m/s
Therefore, by conservation of momentum, we have that the momentum of bullet before collision as follows:
Pball + Pbullet
⇒ 0.01 kg × u = 5 kg m/s
or ⇒ u =

5

0.01

= 500 m/s