AP EAMCET · PHYSICS · Center of Mass Momentum and Collision
A bullet of mass \(30 \mathrm{~g}\) moving with \(700 \mathrm{~ms}^{-1}\) collides with a block of mass \(4 \mathrm{~kg}\) hanging by a string of length \(0.4 \mathrm{~m}\). After collision, the block rises to a height of \(0.2 \mathrm{~m}\). Then, find the velocity of the bullet when it comes out of the block.
- A \(200 \mathrm{~ms}^{-1}\)
- B \(433 \mathrm{~ms}^{-1}\)
- C \(400 \mathrm{~ms}^{-1}\)
- D \(332 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(433 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, mass of bullet, \(m_b=30 \mathrm{~g}=0.03 \mathrm{~kg}\) Velocity of bullet, \(v_b=700 \mathrm{~ms}^{-1}\) Mass of block, \(m_B=4 \mathrm{~kg}\) Height upto which block rises, h = 0. 2m Before collision, momentum of bullet \(=m_b \times v_b\)…
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