AP EAMCET · PHYSICS · Center of Mass Momentum and Collision
. A bullet of mass \(0.01 \mathrm{~kg}\) travelling at a speed of \(500 \mathrm{~ms}^{-1}\) strikes a block of mass \(2 \mathrm{~kg}\) which is suspended by a string of length \(5 \mathrm{~m}\). The centre of gravity of the block is found to rise a vertical distance of \(0.1 \mathrm{~m}\). What is the speed of the bullet after it emerges from the block?
- A \(200 \mathrm{~ms}^{-1}\)
- B \(220 \mathrm{~ms}^{-1}\)
- C \(204 \mathrm{~ms}^{-1}\)
- D \(284 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(220 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
According to question, the given situation is shown in the following figure. Mass of block, \(M=2 \mathrm{~kg}\) Mass of bullet, \(m_1=0.01 \mathrm{~kg}\) Speed of buIlet, \(v=500 \mathrm{~ms}^{-1}\) \[ h=0.1 \mathrm{~m} \] If \(v_1\) and \(v_2\) be the velocities of the bullet…
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