Answer:
(a) When dropped from same height, the two masses
hit the ground with same velocity=\[\sqrt{2gh}\]
\[\therefore \] \[\frac{{{p}_{2}}}{{{p}_{1}}}=\frac{{{m}_{2}}}{{{m}_{1}}}=n\]
(b) We know that K.E. \[K=\frac{{{p}^{2}}}{2m}\] \[\therefore
\]\[\text{p}=\sqrt{2mK}\]
As K is same, \[\frac{{{p}_{2}}}{{{p}_{1}}}=\sqrt{\frac{{{m}_{2}}}{{{m}_{1}}}}=\sqrt{n}\]
You need to login to perform this action.
You will be redirected in
3 sec