Question

When a shell of mass 10kg is fired from a cannon ,a force of 20,000N is exerted on the shell for 0.1 s. What is the muzzle velocity of the shell?

Answer 1

We will be using Newton's Second Law of Motion.

Newton's Second Law states:

The external force applied on an object is equal to the rate of change of momentum of the object with respect to time.

What it simply says is that:

$\sf F = \dfrac{\Delta p}{\Delta t} = \dfrac{\textsf{Change in Momentum}}{\textsf{Time Interval}}$

Also, Momentum is the product of mass and velocity.

$\sf p = mv = \textsf{Mass \times velocity}$

We have here some values, using which an unknown value is to be found. Let us write our data:

Force = F = 20,000 N

Mass = m = 10 kg

Time Interval = 0.1 s

So we have:

$\sf \displaystyle F=\frac{\Delta p}{\Delta t}\\\\\\\implies 20000=\frac{\Delta p}{0.1}\\\\\\\implies \Delta p = 20000\times 0.1\\\\\\\implies \Delta p = 2000\ kg\ m/s\\\\\\\implies \Delta(mv)=2000\\\\\\\implies m\times\Delta v=2000\\\\\\\implies 10\times \Delta v=2000\\\\\\\implies \Delta v=\frac{2000}{10}\\\\\\\implies \Delta v=200\ kg\ m/s$

Thus, Change in velocity of the fired shell is 200 m/s.

Muzzle Velocity is the velocity acquired by the shell as it is fired by the cannon.

Since the shell was initially stationary inside the cannon, its initial velocity was zero.

So,

$\sf \Delta v=\textsf{Final velocity - Initial Velocity}\\\\\\\implies 200=v-0\\\\\\\implies \huge \boxed{\bold{v=200\ m/s}}$

Thus, Muzzle Velocity of the shell is 200 m/s.