Question

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 16.7 kg and an initial velocity of = 7.26 m/s, due east. Object B, however, has a mass of mB = 29.3 kg and an initial velocity of = 4.39 m/s, due north. Find the (a) magnitude and (b) direction of the total momentum of the two-object system after the collision.

Answer 1

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 = $\frac{m_a}{m_a + m_b} \ v_{ao}$

let's calculate

          vₓ = $\frac{16.7}{16.7 + 29.3} \ 7.26$

          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 = $\frac{m_b}{m_a+m_b} \ v_{bo}$

let's calculate

          v_y = $\frac{29.3}{16.7 + 29.3} \ 4.39$

          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 = $\sqrt{v_x^2 + v_y^2}$

      v = $\sqrt{2.636^2 + 2.796^2}$

      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