Questions: A 3050 kg cannon is mounted so that it can recoil only in the horizontal direction. (a) Calculate its recoil velocity in m / s when it fires a 15.0 kg shell at 480 m / s at an angle of 20.0° above the horizontal. (Enter the magnitude.) 2.217 m / s (b) What is the kinetic energy of the cannon in J? 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? Transferred to the ground

Solution Steps

We need to calculate the recoil velocity of a cannon when it fires a shell. The cannon can only recoil horizontally. The shell is fired at an angle, so we need to consider the horizontal component of the shell's velocity to find the recoil velocity of the cannon.

The law of conservation of momentum states that the total momentum before and after firing must be equal. Initially, the system (cannon + shell) is at rest, so the total initial momentum is zero. After firing, the momentum of the cannon and the shell must cancel each other out.

The horizontal component of the shell's velocity is given by: $$v_{\text{shell, horizontal}} = v_{\text{shell}} \cdot \cos(\theta) = 480 \, \text{m/s} \cdot \cos(20^\circ)$$

The horizontal momentum of the shell is: $$p_{\text{shell, horizontal}} = m_{\text{shell}} \cdot v_{\text{shell, horizontal}}$$

The recoil velocity of the cannon $v_{\text{cannon}}$ can be found using: $$m_{\text{cannon}} \cdot v_{\text{cannon}} = -p_{\text{shell, horizontal}}$$

Substitute the known values: $$v_{\text{shell, horizontal}} = 480 \cdot \cos(20^\circ) \approx 450.532 \, \text{m/s}$$ $$p_{\text{shell, horizontal}} = 15.0 \, \text{kg} \cdot 450.532 \, \text{m/s} \approx 6757.98 \, \text{kg} \cdot \text{m/s}$$ $$3050 \, \text{kg} \cdot v_{\text{cannon}} = -6757.98 \, \text{kg} \cdot \text{m/s}$$ $$v_{\text{cannon}} = -\frac{6757.98}{3050} \approx -2.2157 \, \text{m/s}$$

The magnitude of the recoil velocity is $2.2157 \, \text{m/s}$.

The kinetic energy $K$ of the cannon is given by: $$K = \frac{1}{2} m_{\text{cannon}} v_{\text{cannon}}^2$$ $$K = \frac{1}{2} \cdot 3050 \cdot (2.2157)^2 \approx 7480.5 \, \text{J}$$

The vertical component of momentum is transferred to the ground, as the cannon can only recoil horizontally.

Final Answer