Question

Consider an elastic collision between two pucks on an air-hockey table (no friction). The first

puck has a mass of 0.500 kg and is travelling at 4.00 m/s along the x-axis. The second puck
has a mass of 0.300 kg and is at rest. After the collision, the first puck has a velocity of 2.00 m/s
in an unknown direction, O, and the second puck travels at an unknown velocity v2₂ and an
unknown direction of O₂. Determine the 3 missing values.
(6 marks)

Answer 1

To solve this problem, we can use the conservation of momentum and the conservation of kinetic energy.

Conservation of momentum:
m1v1i + m2v2i = m1v1f + m2v2f

where
m1 = 0.500 kg (mass of first puck)
v1i = 4.00 m/s (initial velocity of first puck)
m2 = 0.300 kg (mass of second puck)
v2i = 0 m/s (initial velocity of second puck)
v1f = 2.00 m/s (final velocity of first puck)
v2f = v2₂ (final velocity of second puck)

Substituting in the values, we get:
(0.500 kg)(4.00 m/s) + (0.300 kg)(0 m/s) = (0.500 kg)(2.00 m/s) + (0.300 kg)(v2₂)

Simplifying and solving for v2₂, we get:
v2₂ = 6.00 m/s

Now we can use the conservation of kinetic energy:
(1/2)m1v1i^2 + (1/2)m2v2i^2 = (1/2)m1v1f^2 + (1/2)m2v2f^2

Substituting in the values, we get:
(1/2)(0.500 kg)(4.00 m/s)^2 + (1/2)(0.300 kg)(0 m/s)^2 = (1/2)(0.500 kg)(2.00 m/s)^2 + (1/2)(0.300 kg)(6.00 m/s)^2

Simplifying and solving for v2f, we get:
v2f = 2.00 m/s

Therefore, the three missing values are:
- The final velocity of the second puck is 6.00 m/s in an unknown direction.
- The direction of the final velocity of the first puck is unknown.
- The final velocity of the second puck is 2.00 m/s in the opposite direction of the initial velocity of the first puck.