# Q. A cannon of mass I 000 kg, located at the base of an inclined plane fires a shelI of mass 100 kg in a horizontal direction with a velocity 180 $kmh^{-1}$. 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 metre, to which the cannon ascends the inclined plane as a result of the recoiI is (g = 10 $ms^{-1})$

EAMCETEAMCET 2004Laws Of Motion

Solution:

## Given, Mass of shell $m_1$ = 100 kg Mass of cannon $m_2$ = 1000 kg Velocity of shell = u = 180 $kmh^{-1}$ $\, \, \, \, \, \, \, \, \, = \frac{180 \times 5}{18} = 50 ms^{-1}$ Let velocity of cannon be v. Hence, recoil velocity of cannon from conservation of momentum is. $\, \, \, \, \, \, \, \, \, m_2 v = m_1 u$ $\, \, \, \, \, \, v = \frac{m_1u}{m_2} = \frac{100 \times 50}{1000} = 5 ms^{-1}$ The relation for the acceleration along inclined plane is $\, \, \, \, \, \, \, \, a = g (sin \, \theta + \mu \, cos \, \theta)$ $\, \, \, \, \, \, \, \, \, \, = 10 (sin \, 45^\circ + 0.5 \times cos \, 45^\circ)$ $\, \, \, \, \, \, \, \, 10 \bigg( \frac{1}{\sqrt{2}} + 0.5 \times \frac{1}{\sqrt{2}} \bigg)$ $\, \, \, \, \, \, \, \, = \frac{10 \times 1.5}{\sqrt{2}} = \frac{15}{\sqrt{2}}$ The height to which the cannon ascends the inclined as a results of recoil, is given by $\, \, \, \, \, \, \, \, \, v^2 = u^2 + 2as \, \, \, \, \, \, \, \, \, \, \, (here \, u = 0)$ $\, \, \, \, \, \, s = \frac{v^2}{2a} = \frac{(5)^2}{2 \times \frac{15}{ \sqrt{2}}} = \frac{5 \sqrt{2}}{6} m \simeq \frac{7}{6}$

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