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JIPMER 2011
Exam
JIPMER
Year
2011
Subject
Physics
Chapter
Laws of Motion

JIPMER 2011 · Physics · Laws of Motion

A bullet of mass 20 g and moving with 600 ~m / s collides with a...

A bullet of mass 20 g and moving with 600 m/s600 \mathrm{~m} / \mathrm{s} collides with a block of mass 4 kg hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 m after collision?

Options

  1. A200 m/s200 \mathrm{~m} / \mathrm{s}
  2. B150 m/s150 \mathrm{~m} / \mathrm{s}
  3. C400 m/s400 \mathrm{~m} / \mathrm{s}
  4. D300 m/s300 \mathrm{~m} / \mathrm{s}

Correct answer

A. 200 m/s200 \mathrm{~m} / \mathrm{s}

Step-by-step solution

According to conservation of linear momentum, m1v1=m1v+m2v2m_1 v_1=m_1 v+m_2 v_2 where v1v_1 is velocity of bullet before collision, vv is velocity of bullet after the collision and v2v_2 is the velocity of block. ∴0.02×600=0.02v+4v2\therefore \quad 0.02 \times 600=0.02 v+4 v_2 Here, v2=2gh=2×10×0.2=2 m/s\quad v_2=\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 0.2}=2 \mathrm{~m} / \mathrm{s} ∴0.02×600=0.02v+4×2\therefore \quad 0.02 \times 600=0.02 v+4 \times 2 ⇒0.02v=12−8\Rightarrow \quad 0.02 v=12-8 ⇒v=40.02=200 m/s\Rightarrow \quad v=\frac{4}{0.02}=200 \mathrm{~m} / \mathrm{s}
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