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

Approximately
`2.4 * 10^(4)\; {\rm N}`.

Step-by-step 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 = (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 &= (v^(2) - u^(2))/(2\, x) \\ &= (150^(2) - 0^(2))/(2\, (2.6))\; {\rm m\cdot s^(-2)} \\ &\approx 4.33 * 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 * 10^(3))\; {\rm N} \\ &\approx 2.4 * 10^(4)\; {\rm N}\end{aligned}`.