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
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Collision is inelastic. Hence,
$ V_2^{’}-V_1^{’}=V_1-V_2 $
or $ V_2^{’}-\frac{2}{3}=-6-4 $
$ PI_2=\frac{3}{2}\times 20=30\ \text{cm}$
$ V_2^{’}=-10+\frac{2}{3}=-\frac{28}{3}m/s $
or velocity of B is $ V^{2}=V_1^{2}+V_2^{2}+2V_1V_2\cos \theta $ towards left.
BETA
