RECOIL PHYSICS PROBLEM-please help explain how to get to the answer:)

Question

A 2500-kg cannon suddenly fires a 35-kg cannonball at 280 m/s. What is the recoil speed of the cannon? Assume that frictional forces are negligible, and the cannon is fired horizontally. (answer: 3.9 m/s)

Answer 1

This problem involves the conservation of momentum. The total momentum before the cannonball is fired must be equal to the total momentum after the cannonball is fired.

The momentum of the cannonball before it's fired is zero because it's at rest.

The momentum of the cannonball after it's fired is given by:

\(P_{\text{cannonball}} = m_{\text{cannonball}} \cdot v_{\text{cannonball}}\)

where:

\(m_{\text{cannonball}} = 35 \, \text{kg}\) (mass of the cannonball)

\(v_{\text{cannonball}} = 280 \, \text{m/s}\) (velocity of the cannonball).

Now, you have the momentum of the cannonball:

\(P_{\text{cannonball}} = 35 \, \text{kg} \cdot 280 \, \text{m/s} = 9800 \, \text{Ns}\).

Since momentum is conserved, the total momentum of the cannon and cannonball after the firing must equal the total momentum before firing, which is zero.

So:

\(P_{\text{total\_after}} = P_{\text{cannon}} + P_{\text{cannonball}}\)

The cannon and cannonball are initially at rest, so:

\(P_{\text{total\_before}} = 0 \, \text{Ns}\).

So:

\(0 = P_{\text{cannon}} + 9800 \, \text{Ns}\)

Now, you need to find the recoil speed of the cannon (\(v_{\text{cannon\_recoil}}\)):

\(P_{\text{cannon\_recoil}} = -9800 \, \text{Ns}\) (the negative sign indicates opposite direction).

Using the formula for momentum:

\(P_{\text{cannon\_recoil}} = m_{\text{cannon}} \cdot v_{\text{cannon\_recoil}}\)

You can rearrange this to find the recoil speed:

\(v_{\text{cannon\_recoil}} = \frac{P_{\text{cannon\_recoil}}}{m_{\text{cannon}}\)

Now you can calculate the recoil speed of the cannon:

\(v_{\text{cannon\_recoil}} = \frac{-9800 \, \text{Ns}}{2500 \, \text{kg}} = -3.92 \, \text{m/s}\).

Since speed is a scalar, the recoil speed is \(3.92 \, \text{m/s}\) in the opposite direction of the cannonball. The negative sign indicates that the direction is opposite to that of the cannonball, as expected.

So, the recoil speed of the cannon is approximately \(3.92 \, \text{m/s}\).

I hope this helps out, send a message to me any question