Problem 47PE

A 3000-kg cannon is mounted so that it can recoil only in the horizontal direction. (a) Calculate its recoil velocity when it fires a 15.0-kg shell at 480 m/s at an angle of 20.0º above the horizontal. (b) What is the kinetic energy of the cannon? This energy is dissipated as heat transfer in shock absorbers that stop its recoil. (c) What happens to the vertical component of momentum that is imparted to the cannon when it is fired?

Solution to 47PE

Step 1 of 5</p>

We need to calculate the recoil velocity of a 3000kg canon which can recoil only on horizontal direction. We have to find the kinetic energy of the canon and we have to comment on the vertical component of momentum when the shell is fired.

According to the conservation of momentum, the net change in momentum is zero,

is the initial momentum of canon

is the initial momentum of shell

final momentum of canon

final momentum of shell.

Free body diagram:

Step 2 of 5</p>

Given data:

Mass of the canon

Mass of the shell

Velocity of canon

Angle in which the shell is fired,

Velocity of the shell initially

Recoil velocity

Velocity of the shell finally,

Step 3 of 5</p>We need to find the recoil velocity of the canon

From the free body diagram, we need to resolve velocity vector of the shell.

Horizontal component,

Vertical component,

Applying the conservation of momentum and substitute the given values,

For horizontal direction,

Thus the recoil velocity of the canon is given as