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Sagot :
Answer:
a) v = 3,843 m / s, b) 46.7º North- East
Explanation:
Moment is a vector quantity, so one of the best ways to solve this problem is to solve each component separately.
The system is formed by the two vehicles so that the moment is preserved during the crash
Direction to the East
initial instant. Before the crash
p₀ = mₐ vₐ₀
final insttne. After the crash
p_f = (mₐ + m_b) vₓ
p₀ = p_f
mₐ vₐ₀ = (mₐ + m_b) vₓ
vₓ = [tex]\frac{m_a}{m_a + m_b} \ v_{ao}[/tex]
let's calculate
vₓ = [tex]\frac{16.7}{16.7 + 29.3} \ 7.26[/tex]
vₓ = 2,636 m / s
direction north
initial p₀ = m_b v_{bo}
final p_f = (mₐ + m_b) v_y
p₀ = p_f
m_b v_{bo} = (mₐ + m_b) v_y
v_y = [tex]\frac{m_b}{m_a+m_b} \ v_{bo}[/tex]
let's calculate
v_y = [tex]\frac{29.3}{16.7 + 29.3} \ 4.39[/tex]
v_y = 2.796 m / s
the final speed of the two two vehicles is
v = (2,636 i ^ + 2,796 j ^) m / s
a) the magnitude of the velocity
let's use the Pythagorean theorem
v = [tex]\sqrt{v_x^2 + v_y^2}[/tex]
v = [tex]\sqrt{2.636^2 + 2.796^2}[/tex]
v = 3,843 m / s
b) let's use trigonometry to find the direction
tan θ = v_y / vₓ
θ = tan⁻¹ v_y / vₓ
θ = tan⁻¹ (2,796 / 2,636)
θ = 46.7º
This direction is 46.7º North East
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