A\(3000\;{\rm{kg}}\)cannon is mounted so that it can recoil only in the horizontal direction. (a) Calculate its recoil velocity when it fires a\(15.0\;{\rm{kg}}\)shell at\(480\;{\rm{m/s}}\)at an angle of\(20.0^\circ \)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?

The amount of momentum that an object possesses is determined by two factors: the amount of material moving and the speed at which it is travelling. The factors mass and velocity influence momentum. The momentum of an object is equal to the mass times the velocity of the object in terms of an equation.

The mass of the cannon is\({{m}_{1}}{ = 3000}\;{kg}\)

The mass of the shell is\({m_2} = 15\;{\rm{kg}}\) .

The speed of the shell before firing is\({{v}_{2}}{ = 480}\;{m/s}\).

The angle of the shell with the horizontal is\(\theta = 20.0^\circ \).

a. Using the conservation of momentum along horizontal direction we get,

\(\begin{array}{c}{m_2}{v_2}\cos \theta + {m_1}{v_1} = 0\\{v_1} = \frac{{ - {m_2}{v_2}\cos \theta }}{{{m_1}}}\end{array}\)

Substituting the values we get,

\(\begin{array}{c}{v_1} = \frac{{ - 15 \times 480 \times \cos 20^\circ }}{{3000}}\;{\rm{m/s}}\\ = - 2.26\;{\rm{m/s}}\end{array}\)

b. The kinetic energy of the cannon is,

\(\begin{aligned}{}KE &= \frac{1}{2}{m_1}{v_1}^2\\ &= \frac{1}{2} \times 3000 \times {\left( {2.26} \right)^2}\;{\rm{J}}\\ &= 7660\;{\rm{J}}\end{aligned}\)