Question

A 80.0 kg ice hockey goalie, originally at rest, catches a 0.150 kg hockey puck slapped at him at a velocity of 18.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities (in m/s) be in this case? (Assume the original direction of the ice puck toward the goalie is in the positive direction. Indicate the direction with the sign of your answer.)

Answer 1

The total momentum of the system is a conserved quantity. Equating the total momentum before and after the collision for objects with mass mₐ and mᵦ

mₐ vᵢₐ + mᵦ vᵢᵦ = mₐ vfₐ + mᵦ vfᵦ ----------> [i and f denoting initial and final velocities]

or

initial momentum = final momentum

This equation is valid for any 1-dimensional collision.

Note that, assuming we know the masses of the colliding objects, the above equation only fully describes the collision given the initial velocities of both objects, and the final velocity of at least one of the objects.

An elastic collision is one in which the total kinetic energy of the two colliding objects is the same before and after the collision. For an elastic collision, kinetic energy is conserved. That is:

0.5 mₐ vᵢₐ² + 0.5 mᵦ vᵢᵦ² = 0.5 mₐ vfₐ² + 0.5 mᵦ vfᵦ²

The collision is fully specified given the two initial velocities and masses of the colliding objects.

Combining the above equations gives a solution to the final velocities for an elastic collision of two objects:

vfₐ = [(mₐ - mᵦ) vᵢₐ + 2 mᵦ vᵢᵦ]/[mₐ + mᵦ]

vfᵦ = [2 mₐ vᵢₐ − (mₐ - mᵦ) vᵢᵦ]/[mₐ + mᵦ]

substituting the values

mₐ = 80 kg
mᵦ = 0.150 kg
vᵢₐ = 0 m/s
vᵢᵦ = 18 m/s

=> vfₐ = [(80 - 0.150) 0 + 2(0.150)18]/[80.150]

= 0.06737 m/s ------------------->(in the original direction of the puck)

and

=> vfᵦ = [0 − (80 - 0.150)18]/[80.150] = -17.9326 m/s (opposite direction)