Work Energy And Power Question 317

Question: Two blocks A and B of masses m and 2m placed on a smooth surface are travelling in opposite directions with velocities of 6 m/s and 4 m/s respectively. A perfectly elastic spring is attached to block A. If after collision, velocity of A is $ \frac{2}{3}m/s $ towards right , then velocity of block B would be

Options:

A) $ \frac{4}{3}m/s $ towards left

B) $ \frac{16}{3}m/s $ towards left

C) $ \frac{28}{3}m/s $ m/s towards left

D) 4 m/s towards left

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Answer:

Correct Answer: C

Solution:

[c] Collision is elastic. Hence,

$ V_2^{’}-V_1^{’}=V_1-V_2 $

or $ V_2^{’}-\frac{2}{3}=-6-4 $

$ PI_2=\frac{3}{2}\times 20=30cm $

$ V_2^{’}=-10+\frac{2}{3}=-\frac{28}{3}m/s $

or velocity of B is $ P^{2}=P_1^{2}+P_2^{2}+2P_1P_2\cos \theta $ towards left.



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