Question

A 52.07kg cannon has a recoil velocity of 7.75m/s after it launches a 9.06kg projectile horizontally. The cannon ball hits a 26.28kg target (haystack) several meters away. It takes only 0.32 seconds for the cannon ball to pass through the target breaking it into two segments of equal mass one launching at 27.89 degrees off center in the x, the other 39.62 degrees off center in the y. Assume the cannon ball continues in a true horizontal trajectory.what is the momentum of the cannon ball before the collision with the haystack?

Answer 1

Given data:

* The mass of the cannon is 52.07 kg.

* The recoil velocity of the cannon is 7.75 m/s.

* The mass of the cannon ball is 9.06 kg.

Solution:

By the law of conservation of the momentum,

The momentum of the cannon ball before the collision is the same as the momentum of the cannon ball from the cannon.

The momentum of the cannon ball from the cannon is,

`p=mv`

where m is the mass of the cannon and v is the recoil velocity of the cannon,

Substituting the known values,

`\begin{gathered} p=52.07*7.75 \\ p=403.54kgms^(-1) \end{gathered}`

Thus, the momentum of the cannon ball before the collision is 403.54 kgm/s.