Respuesta :
Answer:
[tex]63.4[/tex] deg
[tex]19.94[/tex] m
Explanation:
[tex]M[/tex] = mass of the heavier truck = 3 tonnes
[tex]V_{i}[/tex] = velocity of truck before collision = [tex]0 \hat{i} + 60 \hat{j}[/tex] kmh⁻¹
[tex]m[/tex] = mass of lighter truck = 1 tonne
[tex]v_{i}[/tex] = velocity of lighter truck before collision = [tex]90 \hat{i} + 0 \hat{j}[/tex] kmh⁻¹
[tex]V_{f}[/tex] = velocity of combination after collision
Using conservation of momentum
[tex]MV_{i}+ m v_{i} = (M + m)V_{f}[/tex]
[tex](3)(0\hat{i} + 60\hat{j}) + (1)(90\hat{i} + 0\hat{j})= (3 + 1) V_{f}[/tex]
[tex]V_{f} =22.5\hat{i} + 45\hat{j}[/tex]
magnitude of the final speed is given as
[tex]|V_{f}| = \sqrt{22.5^{2}+45^{2}} = 50.31[/tex] km/h
Direction is given as
[tex]\theta = tan^{-1}\left ( \frac{45}{22.5} \right ) = 63.4[/tex] deg
[tex]V_{f}[/tex] = velocity of combination after collision = 50.31 km/h = 13.98 m/s
[tex]V[/tex] = velocity of combination after it comes to stop = 0 m/s
[tex]a[/tex] = acceleration = [tex]\mu _{k} g[/tex] = - [tex](0.5) (9.8)[/tex] = - 4.9 m/s²
[tex]d[/tex] = stopping distance
using the equation
[tex]V^{2} = V^{2}_{f} + 2 ad\\\\(0)^{2} = (13.98)^{2} + 2 (- 4.9)d\\[/tex]
[tex]d = 19.94[/tex] m
