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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Advanced 2011 · Paper 2 · Physics · Center of Mass, Momentum and Collision

A ball of mass 0.2 ~kg rests on a vertical post of height 5 ~m . A...

A ball of mass 0.2 kg0.2 \mathrm{~kg} rests on a vertical post of height 5 m5 \mathrm{~m}. A bullet of mass 0.01 kg0.01 \mathrm{~kg}, travelling with a velocity v m/sv \mathrm{~m} / \mathrm{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 m20 \mathrm{~m} and the bullet at a distance of 100 m100 \mathrm{~m} from the foot of the post. The initial velocity vv of the bullet is 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

Options

  1. A250 m/s250 \mathrm{~m} / \mathrm{s}
  2. B2502 m/s250 \sqrt{2} \mathrm{~m} / \mathrm{s}
  3. C400 m/s400 \mathrm{~m} / \mathrm{s}
  4. D500 m/s500 \mathrm{~m} / \mathrm{s}

Correct answer

D. 500 m/s500 \mathrm{~m} / \mathrm{s}

Step-by-step solution

Time taken by the bullet and ball to strike the ground is t=2hg=2×510=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1  sLet\mathrm{~s}Letv_1andandv_2arethevelocitiesofballandbulletaftercollision.ThenapplyingWehave,are the velocities of ball and bullet after collision. Then applying We have,20=v_1 \times 1ororv_1=20 \mathrm{~ms}^{-1}100=v2100=v_2\times11\text{ or }v2=100v_2=100\mathrm{~m}/\mathrms−1Now,fromconservationoflinearmomentumbeforeandaftercollisionwehave,/ \mathrms^{-1} Now, from conservation of linear momentum before and after collision we have,0.01 v=(0.2 \times 20)+(0.01 \times 100)Onsolving,wegetv=500On solving, we get v=500\mathrm{~ms}^{-1}\therefore Correct answer is (d). Analysis of Question Question is moderately lengthy from calculation point of view, otherwise it is simple.
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