Day 2: Let It Snow!

Glossary: $\text{Hill height} = H \\ \text{Snow mass} = m \\ \text{Tommy mass} = M \\ \text{Hill conversion efficiency} = \epsilon \\ \text{Snow friction coefficient} = \mu \\ \text{Gravity acceleration} = g \\$

Compute these steps sequentially to get the final answer:

Change in gravitational potential energy: $\Delta U = M g H$

Kinetic energy at hill bottom: $E_b = \epsilon \, \Delta U$

Speed at bottom: $v_b = \sqrt{\frac{2 E_b}{M}}$

Momentum conservation upon collision with snowball: $M \, v_b = (M + m) \, v_f \\ v_f = \frac{M \, v_b}{M + m}$

Kinetic energy of combined mass after collision: $E_f = \frac{1}{2} (M + m) \, v_f^2$

Friction force on combined mass: $F = \mu \, (M + m) \, g$

Distance slid over snow: $E_f = F \, D \\ D = \frac{E_f}{F}$

Computing all of this results in $D \approx 8.49$

We know that Tommy's energy at the top of the hill is $E_i \ = \ mgh \ = \ (55)(9.81)(20) \ = \ 10791$ . We know that all of the gravitational potential energy will be converted to kinetic energy, with $30$ % efficiency, so the energy at the bottom of the hill will be $(E_i)(PE) \ = \ (10791)(0.3) \ = \ 3237.3$ . Solving for the velocity, we get $E_k \ = \ \frac{1}{2}mv^2 \ = \ 3237.3 \ \Rightarrow \ v \ = \ 10.8 \ m/s$ . Tommy will then slide across the lake at the same velocity and collide with the snowball. Because of conservation of momentum, we get $\Sigma \ p_i \ = \ \Sigma \ p_f \ \Rightarrow \ (55)(10.8) \ = \ (75)v_i$ . So the initial velocity of the combine mass is $v_i \ = \ 7.92 \ m/s$ . We can solve for the acceleration of the combined mass by noting the only unbalanced force acting on the combined mass is friction, which will act in the opposite direction of motion with magnitude $F_f \ = \ mg\mu_k \ = \ (75)(9.81)(0.38) \ = \ 280 \ N$ , so the acceleration is $\frac{F_f}{m}$ , or $-3.73 \ m/s$ . Using the equation $v_f^{2} \ = \ v_i^{2} \ + \ 2ad$ , we can solve for $d$ easily by noting that the final velocity will be $0$ . We get: $-(7.92)^2 \ = \ 2(-3.73)(d) \ \Rightarrow \ d \ = \ 8.41$ .