Explanation:
Part a)
By the conservation of energy, we can write the following equation
[tex]\begin{gathered} U_i-W=K_f \\ mgh-W=\frac{1}{2}mv_f^2 \end{gathered}[/tex]Where m is mass, g is the gravity, h is the height, W is the work done by the force of air resistance, and Vf is the speed at the bottom of the hill.
Replacing m = 85 kg, g = 9.8 m/s², h = 10 m, W = 5 x 10³ J, and solving for Vf, we get
[tex]\begin{gathered} (85\text{ kg\rparen\lparen9.8 m/s}^2)(10\text{ m\rparen-\lparen5}\times10^3J)=\frac{1}{2}(85\text{ kg\rparen v}_f^2 \\ 8330J-5000J=(42.5\text{ kg\rparen v}_f^2 \\ 3330J=(42.5\text{ kg\rparen v}_f^2 \\ \\ \frac{3330\text{ J}}{42.5\text{ kg}}=v_f^2 \\ \\ 78.35=v_f^2 \\ \sqrt{78.35}=v_f \\ 8.85\text{ m/s = v}_f \end{gathered}[/tex]Therefore, the speed at the bottom of the hill is 8.85 m/s
Part b)
We can represent the situation with the following figure
Then, by the conservation of momentum on the x and y-direcction, we can write the following equation
[tex]\begin{gathered} p_{ix}=p_{fx} \\ p_{iy}=p_{fy} \\ \\ m_1v_{1x}=(m_1+m_2)v_{fx}\rightarrow v_{fx}=\frac{m_1v_{1x}}{m_1+m_2} \\ \\ m_2v_{2y}=(m_1+m_2)v_{fy}\rightarrow v_{fy}=\frac{m_2v_{2y}}{m_1+m_2} \end{gathered}[/tex]So, we can calculate the final speed on each direction replacing m1 = 85 kg, v1x = 8.85 m/s, m2 = 52 kg, v2y = 4.2 m/s as follows
[tex]\begin{gathered} v_{fx}=\frac{(85\text{ kg\rparen\lparen8.85 m/s\rparen}}{85\text{ kg + 52 kg}}=5.49\text{ m/s} \\ \\ v_{fy}=\frac{(52\text{ kg\rparen\lparen4.2 m/s\rparen}}{85\text{ kg + 52 kg}}=1.59\text{ m/s} \end{gathered}[/tex]Then, by the Pythagorean theorem, we get that the speed after the collision is
[tex]\begin{gathered} v_f=\sqrt{(5.49\text{ m/s\rparen}^2+(1.59\text{ m/s\rparen}^2} \\ v_f=5.72\text{ m/s} \end{gathered}[/tex]And the direction of the speed is
[tex]\begin{gathered} \theta=\tan^{-1}(\frac{v_{fy}}{v_{fx}}) \\ \\ \theta=\tan^{-1}(\frac{1.59}{5.49}) \\ \\ \theta=16.15\degree \end{gathered}[/tex]So, the answer is
The speed is 5.72 m/s at a direction of 16.15°
