Question

A\(90.0\;{\rm{kg}}\)ice hockey player hits a\(0.150\;{\rm{kg}}\)puck, giving the puck a velocity of\(45.0\;{\rm{m/s}}\). If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal\(15.0\;{\rm{m}}\)away?

Short answer

The distance that the player recoils is\(0.025\;{\rm{m}}\)

Step-by-step solution

Definition of Final Velocity

A final velocity is defined as the final speed of a moving object with an initial velocity and acceleration over some time.

The mass of the player is\({m_1} = 90.0\;{\rm{kg}}\).

The mass of the puck is\({m_2} = 0.150\;{\rm{kg}}\).

The final velocity of the puck is\({v_2} = 45.0\;{\rm{m/s}}\).

The distance from the puck is\(x = 15\;{\rm{m}}\).

Calculation of final speed of the player

Using the conservation of momentum along vertical direction we get,

\(\begin{array}{c}0 = {m_1}{v_1} + {m_2}{v_2}\\0 = 90.0{v_1} + 0.150 \times 45.0\\{v_1} = - 0.075\;{\rm{m/s}}\end{array}\)

Therefore the final speed of the player is \( - 0.075\;{\rm{m/s}}\)

Calculation of the distance of the player

The time taken by the puck is,

\(\begin{array}{c}t = \frac{x}{{{v_2}}}\\ = \frac{{15.0}}{{45.0}}\;{\rm{s}}\\ = 0.33\;{\rm{s}}\end{array}\)

The distance the player recoils is,

\(\begin{array}{c}d = {v_1}t\\ = 0.075 \times 0.33\;{\rm{m}}\\ = 0.025\;{\rm{m}}\end{array}\)

Therefore the distance the player recoils is \(0.025\;{\rm{m}}\)