TS EAMCET · Physics · Center of Mass Momentum and Collision
A man of \(50 \mathrm{~kg}\) is standing at one end on a boat of length \(25 \mathrm{~m}\) and mass \(200 \mathrm{~kg}\). If he starts running and when he reaches the other end, he has a velocity \(2 \mathrm{~ms}^{-1}\) with respect to the boat. The final velocity of the boat is : (in \(\mathrm{ms}^{-1}\) )
- A \(\frac{2}{5}\)
- B \(\frac{2}{3}\)
- C \(\frac{8}{5}\)
- D \(\frac{8}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{5}\)
Step-by-step Solution
Detailed explanation
Let \(v\) be the velocity of the boat with respect to the water, then from conservation of linear momentum. \((200+50) v+50 \times 2=50 \times 0+200 \times 0\) \(250 v=-100\) \(v=-\frac{100}{250}=-\frac{2}{5} \mathrm{~m} / \mathrm{s}\)
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