| 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 |
 |
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
Correct Answer: C
Solution :
| [c] Collision is elastic. Hence, |
| \[V_{2}^{'}-V_{1}^{'}={{V}_{1}}-{{V}_{2}}\] |
| or \[V_{2}^{'}-\frac{2}{3}=-6-4\] |
| \[P{{I}_{2}}=\frac{3}{2}\times 20=30cm\] \[V_{2}^{'}=-10+\frac{2}{3}=-\frac{28}{3}m/s\] |
| or velocity of B is \[{{P}^{2}}=P_{1}^{2}+P_{2}^{2}+2{{P}_{1}}{{P}_{2}}\cos \theta \] towards left. |
You need to login to perform this action.
You will be redirected in
3 sec
