A particle of mass m, = 1.000 kg moves at speed v_1 = 0.500 m/s. It collides wit

Question

A particle of mass m, = 1.000 kg moves at speed v_1 = 0.500 m/s. It collides with a particle ot mass m_2 = 2.000 kg at rest. What is the total momentum of the system in the x direction before the collision? What is the total momentum of the system in the y direction before the collision? After the collision, m_1 moves with speed v_3 at an angle theta_4 = 315.0 degree with respect to the x axis, and m: moves with speed v_4 at an angle theta_4 = 30.0 degree with respect to the x axis. Write an expression for the total momentum of the system in the x direction and another expression for the total momentum in the y direction after the collision in terms of the symbols m_1. M_2. V_2. V_4, and angles theta_3 and theta_4. Equate the expression for the x component in Question 2 to the value of the x component in Question 1. Equate the expression for they component in Question 2 to the value of they component in Question 1. In the resulting two equations, v_3 and v_4 are the only two unknowns. Solve the two equations for v_3 and v_4. Show work below.

Explanation / Answer

1.
a)
in x direction, momentum = m1*v1 = 1 Kg * 0.5 m/s = 0.5 Kgm/s
b.
in y direction, there is no movement, so momentum = 0

2.
x momentum = m2*v4*cos thetha 4 + m1*v3*cos thetha3
y momentum = m2*v4*sin thetha 4 + m1*v3*sin thetha3

3.
equate momentum in y direction:
m2*v4*sin thetha 4 + m1*v3*sin thetha3=0
2*v4*sin 30 + 1*v3*sin 315 =0
v4 - 0.7071v3=0
so,
v4 = 0.7071v3

equate momentum in x direction:
m2*v4*cos thetha 4 + m1*v3*cos thetha3 = 0.5
put v4= 0.7071v3
2*0.7071v3*cos 30 + 1*v3*cos 315= 0.5
1.225 v3 + v3*0.7071= 0.5
v3=0.26 m/s
v4 = 0.7071v3 =0.7071*0.26=0.183 m/s

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