Question

A cannon of mass 2m located at the base of an inclined plane shoots a shell of mass m in horizontal direction with velocity v₀. The angle of inclination of plane is 45° and the coefficient of friction between the cannon and the plane is 0.5. The height to which cannon ascends the plane as a result of recoil is:

• A. $\frac{v_0^2}{2g}$
• B. $\frac{v_0^2}{12g}$
• C. $\frac{v_0^2}{6g}$
• D. $\frac{v_0^2}{g}$

Answer 1

Answer: (B)

$\frac{v_0^2}{12g}$

Answer 2

1. Conservation of Momentum: The cannon and shell system is initially at rest. When the shell of mass $m$ is fired horizontally with velocity $v_0$, the cannon of mass $2m$ recoils. By conservation of linear momentum in the horizontal direction, the recoil velocity of the cannon, $v_c$, is found to be $m v_0 = (2m) v_c$, which gives $v_c = \frac{v_0}{2}$. This recoil velocity is directed up the inclined plane.

2. Forces on the Cannon: As the cannon moves up the inclined plane, it experiences the following forces opposing its motion:

   • Component of gravity along the plane: $F_g = (2m) g \sin\theta$ (downwards)
   • Friction force: $f = \mu N$, where $N$ is the normal force. The normal force is $N = (2m) g \cos\theta$. Thus, $f = \mu (2m) g \cos\theta$ (downwards, opposing motion).

3. Deceleration: The net force acting down the inclined plane is $F_{net} = F_g + f = (2m) g \sin\theta + \mu (2m) g \cos\theta$. The deceleration $a$ of the cannon is given by $F_{net} = (2m) a$. So, $(2m) a = (2m) g (\sin\theta + \mu \cos\theta)$. $a = g (\sin\theta + \mu \cos\theta)$. Given $\theta = 45^\circ$ and $\mu = 0.5$: $a = g \left(\sin 45^\circ + 0.5 \cos 45^\circ\right) = g \left(\frac{1}{\sqrt{2}} + 0.5 \times \frac{1}{\sqrt{2}}\right) = g \left(\frac{1.5}{\sqrt{2}}\right) = g \frac{3}{2\sqrt{2}}$.

4. Kinematics: The cannon starts with an initial velocity $u = v_c = \frac{v_0}{2}$ up the incline and comes to rest ($v=0$). Using the kinematic equation $v^2 = u^2 + 2as$, where $s$ is the distance traveled up the plane: $0^2 = \left(\frac{v_0}{2}\right)^2 + 2 \left(-g \frac{3}{2\sqrt{2}}\right) s$ $0 = \frac{v_0^2}{4} - g \frac{3}{\sqrt{2}} s$ $s = \frac{v_0^2}{4} \times \frac{\sqrt{2}}{3g} = \frac{\sqrt{2} v_0^2}{12g}$.

5. Height Ascended: The height $h$ to which the cannon ascends is related to the distance $s$ along the plane by $h = s \sin\theta$. $h = \left(\frac{\sqrt{2} v_0^2}{12g}\right) \sin 45^\circ = \left(\frac{\sqrt{2} v_0^2}{12g}\right) \left(\frac{1}{\sqrt{2}}\right) = \frac{v_0^2}{12g}$.