Question

A 5.5 kg cannon ball leaves a canon with a speed of 150 m/s. Find the average net force applied to the ball if the cannon muzzle is 2.6 m long.

Answer 1

Answer:

Approximately $2.4 \times 10^{4}\; {\rm N}$.

Explanation:

Assume that the acceleration $a$ of the cannonball is constant while in the muzzle. Apply the following SUVAT equation to find this acceleration:

$v^{2} - u^{2} = 2\, a\, x$.

$\displaystyle a = \frac{v^{2} - u^{2}}{2\, x}$,

Where:

• $v = 150\; {\rm m\cdot s^{-1}}$ is the velocity after the acceleration,
• $u = 0\; {\rm m\cdot s^{-1}}$ is the initial velocity (assuming the cannonball was initially not moving,) and
• $x = 2.6\; {\rm m}$ is the displacement during the acceleration.

Thus, under the assumptions, acceleration of the cannonball in the muzzle would be:

$\begin{aligned} a &= \frac{v^{2} - u^{2}}{2\, x} \\ &= \frac{150^{2} - 0^{2}}{2\, (2.6)}\; {\rm m\cdot s^{-2}} \\ &\approx 4.33 \times 10^{3}\; {\rm m\cdot s^{-2}}\end{aligned}$.

It is given that the mass of this cannonball is $m = 5.5\; {\rm kg}$. By Newton's Laws of Motion, the net force on the cannonball would be:

$\begin{aligned}F_{\text{net}} &= m\, a \\ &\approx (5.5)\, (4.33 \times 10^{3})\; {\rm N} \\ &\approx 2.4 \times 10^{4}\; {\rm N}\end{aligned}$.