A) \[\frac{7}{6}\]
B) \[\frac{5}{9}\]
C) \[\frac{2}{6}\]
D) \[\frac{1}{6}\]
Correct Answer: A
Solution :
Given: Mass of shell\[{{m}_{1}}=100\,kg\] Mass of cannon \[{{m}_{2}}=1000\,kg\] Velocity of shell \[=u=180\,km/h\] \[=\frac{180\times 5}{18}=50\,m/s\] Let velocity of cannon = v Hence, recoil velocity of cannon from conservation of momentum is \[{{m}_{2}}v={{m}_{1}}u\] \[v=\frac{{{m}_{1}}u}{{{m}_{2}}}=\frac{100\times 50}{1000}=5\,m/s\] The relation for the acceleration along inclined plane is \[a=g(sin\theta +\mu cos\theta )\] \[=10(sin{{45}^{o}}+0.5\times \cos {{45}^{o}})\] \[=10\left( \frac{1}{\sqrt{2}}+0.5\times \frac{1}{\sqrt{2}} \right)\] \[=\frac{10\times 1.5}{\sqrt{2}}=\frac{15}{\sqrt{2}}\] The height to which the cannon ascends the inclined as a results of recoil, is given by \[{{v}^{2}}={{u}^{2}}+2as\] (here \[u=0\]) \[s=\frac{{{v}^{2}}}{2a}=\frac{{{(5)}^{2}}}{2\times \frac{15}{\sqrt{2}}}=\frac{5\sqrt{2}}{6}m\simeq \frac{7}{6}\]
You need to login to perform this action.
You will be redirected in
3 sec
