Question

A 1,107.64 kg cannon shoots a 51.72 kg cannonball across a field with a velocity of 2.72 m/s. What is the recoil velocity of the cannon?

Answer 1

the recoil velocity of the cannon is 0.12 m/s

Explanation

The momentum of the projectile at the instant it was fired is equal to the product of its mass and its velocity (m*v). Then, by conservation of momentum, the cannon must move in the opposite direction with the same momentum

so

conservation of momemtum

$\begin{gathered} m_{cannon}v_{cannon}=m_{cannonball}v_{cannonball} \\ \end{gathered}$

replace

$\begin{gathered} m_{cannon}v_{cannon}=m_{cannonball}v_{cannonball} \\ (1,107.64\operatorname{kg})v_{cannon}=51.72\operatorname{kg}\cdot2.72\frac{m}{s} \\ 1107.64v_{cannon}=140.6784 \\ divide\text{ both sides by }1107.64 \\ \frac{1107.64v_{cannon}}{1107.64}=\frac{140.6784}{1107.64} \\ v_{cannon}=0.12\text{ m/s} \end{gathered}$

therefore,

the recoil velocity of the cannon is 0.12 m/s

I hope this helps you