A potato cannon is used to launch a potato on a frozen lake, as shown in the figure. The mass of the cannon, \(m_{c}\) is \(10.0 \mathrm{~kg},\) and the mass of the potato, \(m_{\mathrm{p}}\), is \(0.850 \mathrm{~kg} .\) The cannon's spring (with spring constant \(\left.k_{\mathrm{c}}=7.06 \cdot 10^{3} \mathrm{~N} / \mathrm{m}\right)\) is compressed \(2.00 \mathrm{~m}\). Prior to launching the potato, the cannon is at rest. The potato leaves the cannon's muzzle moving horizontally to the right at a speed of \(v_{\mathrm{p}}=175 \mathrm{~m} / \mathrm{s}\). Neglect the effects of the potato spinning. Assume there is no friction between the cannon and the lake's ice or between the cannon barrel and the potato. a) What are the direction and magnitude of the cannon's velocity, \(\mathrm{v}_{c}\), after the potato leaves the muzzle? b) What is the total mechanical energy (potential and kinetic) of the potato/cannon system before and after the firing of the potato?

Initially, the system is at rest, so the total linear momentum is zero. After the launch, the cannon moves in the opposite direction of the potato. Let's assume that positive direction is the direction of the potato's velocity. Using the conservation of linear momentum, we can write:$$ m_{p}v_{p} + m_{c}v_{c} = 0 $$Now we can solve for the cannon's velocity, \(v_{c}\).

Use the formula from Step 1 and solve for \(v_c\):$$ v_{c} = -\frac{m_{p}v_{p}}{m_{c}} $$Plug in the given values for the mass of the potato (\(m_p = 0.850 \, kg\)), velocity of the potato (\(v_p = 175 \, m/s\)), and mass of the cannon (\(m_c = 10.0 \, kg\)):$$ v_c = -\frac{(0.850 \, kg)(175 \, m/s)}{10.0 \, kg} = -14.88 \, m/s $$The negative sign indicates that the cannon's velocity is opposite to the direction of the potato's velocity. Therefore, the magnitude and direction of the cannon's velocity are \(14.88 \, m/s\) to the left.

Initially, the only energy in the system is the potential energy stored in the compressed spring (\(U_{c} = \frac{1}{2}k_{c}x^{2}\)), where \(x\) is the compression length. After firing, the system has kinetic energies for both the potato (\(K_{p}\)) and the cannon (\(K_{c}\)). The total mechanical energy for each case will be given by:$$ E_{initial} = U_{c} $$and$$ E_{final} = K_{p} + K_{c} $$We can calculate \(U_{c}\) using the spring constant, \(k_{c} = 7.06*10^3 \, N/m\), and compression length, \(x = 2.00 \, m\):$$ U_{c} = \frac{1}{2}(7.06 * 10^3 \, N/m)(2.00 \, m)^2 = 28.24*10^3 \, J $$Now we can calculate \(K_{p}\) and \(K_{c}\) using the mass and velocity of the potato and the cannon respectively:$$ K_{p} = \frac{1}{2} m_{p} v_{p}^2 = \frac{1}{2}(0.850\, kg)(175 \,m/s)^2 = 12.98*10^3 \, J $$$$ K_{c} = \frac{1}{2} m_{c} v_{c}^2 = \frac{1}{2}(10.0\, kg)(14.88 \,m/s)^2 = 1.11 * 10^3 \, J $$Adding \(K_{p}\) and \(K_{c}\) gives the final total mechanical energy:$$ E_{final} = K_{p} + K_{c} = 12.98*10^3 \, J + 1.11*10^3 \, J = 14.09*10^3 \, J $$Now we have the total mechanical energies before and after firing the potato:$$ E_{initial} = 28.24*10^{3}\, J $$$$ E_{final} = 14.09*10^{3}\, J $$a) Velocity of the cannon: \(v_c = -14.88 \, m/s\) (leftward) b) Total mechanical energy before firing: \(E_{initial} = 28.24*10^{3}\, J\), Total mechanical energy after firing: \(E_{final} = 14.09*10^{3}\, J\)