A 3.0-kg mass moving in the positive x direction with a speed of 10 m/s collides with a 6.0-kg mass initially at rest.After the collision,the speed of the 3.0-kg mass is 8.0 m/s,and its velocity vector makes an angle of 35$\degree$ with the positive x axis.What is the magnitude of the velocity of the 6.0-kg mass after the collision?
A)2.2 m/s
B)2.9 m/s
C)4.2 m/s
D)3.5 m/s
E)4.7 m/s

Question

Quizplus · Accepted Answer

Conservation of momentum: 
(mass 1 x initial velocity 1) + (mass 2 x initial velocity 2) = (mass 1 x final velocity 1) + (mass 2 x final velocity 2)
(3 kg x 10 m/s) + (6 kg x 0 m/s) = (3 kg x 8 m/s) + (6 kg x vf)
Solving for vf, we get vf = 2.67 m/s.
We can use conservation of kinetic energy to find the angle between the velocity vector of the 3 kg mass and the positive x-axis:
(1/2 x mass 1 x initial velocity 1^2) = (1/2 x mass 1 x final velocity 1^2) + (1/2 x mass 2 x final velocity 2^2)
(1/2 x 3 kg x 10^2) = (1/2 x 3 kg x 8^2) + (1/2 x 6 kg x vf^2)
Solving for vf again, we get vf = 4.06 m/s.
The velocity vector of the 6 kg mass must be in the negative x direction to conserve momentum. Therefore, its magnitude is also 2.67 m/s.

A 30-Kg Mass Moving in the Positive X Direction with a a Speed