Question:

A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling 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 distanceof 100 m from the foot of the post The initial velocity v of the bullet is

Updated On: Jun 14, 2022
  • 250 m/s
  • $ 250 \sqrt m / s$
  • 400 m/s.
  • 500 m/s.
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The Correct Option is D

Solution and Explanation

R = $ u \sqrt{ \frac{ 2 h}{ g}} \Rightarrow 20 = v_1 \sqrt{ \frac{ 2 \times 5 }{ 10 }} \, and \, 100 = v_2 \sqrt{ \frac{ 2 \times 5 }{ 10}} $
$\Rightarrow v_1 = 20 m/s, \, v_2 = 100 m/s.$
Applying momentum conservation just before and just after the collision
(0.01) (v) = (0.2) (20) + (0.01) (100)
v = 500 m/s
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System of Particles and Rotational Motion

  1. The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
  2. The distance between the pair of particles in such a body does not replace or alter. Rotational motion can be described as the motion of a rigid body originates in such a manner that all of its particles move in a circle about an axis with a common angular velocity.
  3. The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.