TS EAMCET · Physics · Center of Mass Momentum and Collision
A bullet of mass \(25 \mathrm{~g}\) moves horizontally at a speed of \(250 \mathrm{~m} / \mathrm{s}\) is fired into a wooden block of mass \(1 \mathrm{~kg}\) suspended by a long string. The bullet crosses the block and emerges on the other side. If the centre of the mass of the block rises through a height of \(20 \mathrm{~cm}\). The speed of the bullet as it emerges from the block is (take, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A 300 m/s
- B 220 m/s
- C 150 m/s
- D 170 m/s
Answer & Solution
Correct Answer
(D) 170 m/s
Step-by-step Solution
Detailed explanation
The given situation is shown in the following figure Let \(v_1\) and \(v_2\) are final velocities of bullet and block, respectively, as block rises upto height \(h\). Hence, according to conservation of energy, \(\frac{1}{2} M v_2^2=M g h \Rightarrow v_2=\sqrt{2 g h}\) Given,…
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