A) \[\frac{mv}{2T}\]
B) \[\frac{3mv}{2T}\]
C) \[\frac{3mv}{T}\]
D) None of these
Correct Answer: B
Solution :
[b] The momentum of the bead at A is \[{{\vec{P}}_{i}}=-mv\hat{i}\] The momentum of the bead at B \[{{\vec{P}}_{f}}=(mv/2)i\] Therefore, the magnitude of the change in momentum between A and B is \[\Delta p=\left| {{{\vec{p}}}_{f}}-{{{\vec{p}}}_{i}} \right|=(3/2)mv\] Average force exerted by the bead on the wire is \[{{F}_{av}}=\frac{\Delta p}{\Delta t}=\frac{3mv}{2T}\]You need to login to perform this action.
You will be redirected in
3 sec