Source: High school physics (Chinese)
Problem Sets:
Problem
A cannon cart of mass $M$ fires a cannonball of mass $m$ at an elevation angle $\theta$. The muzzle velocity of the cannonball relative to the ground is $v_0$. The cart is initially at rest.
- Find the velocity of the cart as the cannonball leaves the muzzle.
- Is the pressure the cart exerts on the ground at this moment equal to the total weight of the cart and the cannonball, $(M+m)g$?
[Q1] $V = -\frac{m v_0 \cos\theta}{M}$ [Q2] No, it is greater than the total weight.
[Q1] Consider the cart and cannonball as a system. The horizontal momentum is conserved because there are no external horizontal forces. The initial momentum is zero. Let $V$ be the cart's recoil velocity. The horizontal component of the cannonball's velocity is $v_{0x} = v_0 \cos\theta$.
$$p_{ix} = p_{fx}$$ $$0 = MV + m v_0 \cos\theta$$ $$V = -\frac{m v_0 \cos\theta}{M}$$[Q2] Consider the vertical forces on the cart-cannonball system. At the moment of firing, the cannonball has a vertical acceleration, so its vertical momentum is increasing. This requires an internal upward force from the cannon on the ball. By Newton's third law, the ball exerts an equal and opposite (downward) force on the cannon cart. Therefore, the normal force $N$ from the ground must support the weight $(M+m)g$ plus this additional downward force. Since $N > (M+m)g$, the pressure on the ground is greater than the total weight.