Question

A box A of mass 4.0 kg is moving toward the right at a speed of 30.0 m/s on a frictionless horizontal surface, and another box B of mass 5.0 kg is sitting stationary to the left of box A. Box A collides with Box B.

(a) Suppose they bounce off each other. The collision is elastic. What are the final speeds of the two boxes, and which direction is each going in now?

(b) Suppose they stick together because they both have Velcro on their outer surfaces. What is the final speed of the two boxes? Which direction are they going in?

Answer 1

Mass of box A, m_A = 4 kg
Mass of box B, m_B = 5 kg
Velocity of box A, v = 30 m/s

(a)
Consider v_A and v_B as the velocity of boxes A and B after the collision.
v_A = [(m_A - m_B) / (m_A + m_B)] * v
= [(4 - 5) / (4 + 5)] * 30
= - 3.33 m/s
The negative sign shows that the direction of velocity is leftward
v_A = 3.33 m/s, towards left

v_B = [(2 * m_A) / (m_A + m_B)] * v
= [(2 * 4) / (4 + 5)] * 30
= 26.7 m/s, towards right

(b)
The initial momentum of the system, Pi = m_A * v
The final momentum of the system, Pf = (m_A + m_B) * v'
Where v' is the final velocity of both the boxes.

Using conservation of momentum,
Pi = Pf
m_A * v = (m_A + m_B) * v'
4 * 30 = (4 + 5) * v'
v' = 120 / 9
= 13.3 m/s