Question

A cannon of mass 1000 kg, located at the base of an inclined plane fires a shell of mass 100 kg in a horizontal direction with a velocity 180 km/h. The angle of inclination of the inclined plane with the horizontal is $ 45{}^\circ $ . The coefficient of friction between the cannon and the inclined plane is 0.5. The height, in m, to which the cannon ascends the inclined plane as a result of the recoil is: $ (g=10m/{{s}^{2}}) $

• A. $ \frac{7}{6} $
• B. $ \frac{5}{9} $
• C. $ \frac{2}{6} $
• D. $ \frac{1}{6} $

Answer 1

Answer: (A)

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} $