JEE Advanced | 2015Math miscellaneousHard
Question
A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, traveling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The velocity V of the bullet is 0 20 100V m/s
Options
A.250 m/s
B.250√2 m/s
C.400 m/s
D.500 m/s
Solution
5 = 1/2 (10)t2 ⇒ t = 1 sec
vball = 20 m/s
vbullet = 100 m/s
0.01 V = 0.01 × 100 + 0.2 × 20
v = 100 + 400 = 500 m/s
vball = 20 m/s
vbullet = 100 m/s
0.01 V = 0.01 × 100 + 0.2 × 20
v = 100 + 400 = 500 m/s
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