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)

$\frac{7}{6}$

Answer 2

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 $u=0$ ) $s=\frac{{{v}^{2}}}{2a}=\frac{{{(5)}^{2}}}{2\times \frac{15}{\sqrt{2}}}=\frac{5\sqrt{2}}{6}m\simeq \frac{7}{6}$