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Question
A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 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 m away?
Question by OpenStax is licensed under CC BY 4.0.
Final Answer
$2.5 \textrm{ cm}$ in the opposite direction.
Solution Video

OpenStax College Physics Solution, Chapter 8, Problem 52 (Problems & Exercises) (2:27)

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Video Transcript

This is College Physics Answers with Shaun Dychko. A hockey player and a puck are initially at rest and the player hits the puck with their stick and the puck moves forward with a speed of 45.0 meters per second and the player is therefore gonna move in the other direction in order for momentum to be conserved and we have to figure out what this velocity <i>v 2 prime</i> is gonna be because we'll multiply that velocity by an amount of time that it takes for this puck to cover 15.0 meters until it gets to the goal and then we'll figure out how far this player moves back the other way. So the puck we are told is 0.150 kilograms and the player is 90.0 kilograms; we have positive defined to the right and we'll start off by saying the total momentum after hitting the puck equals the total momentum initially of which there's no momentum initially since they are not moving so we have zero here and we'll solve for <i>v 2 prime</i> by subtracting <i>m 1v 1 prime</i> from both sides and then divide both sides by <i>m 2</i>. So <i>v 2 prime</i> then is negative of 0.150 kilograms times negative 45.0 meters per second— velocity of the puck— divided by 90.0 kilograms— mass of the player— and that is 0.075 meters per second and that's positive to the right. Now the distance that the puck will cover is going to be 15.0 meters— we'll label it with the number 1 because everything else to do with the puck has a number '1' subscript— and we'll say that it's <i>v 1</i> times <i>t</i>. Now I'm not giving the time a subscript '1' because time for the puck traveling and time for the player moving are the same and so we'll solve for this <i>t</i> by dividing both sides by <i>v 1</i> and then switching the sides around and we have <i>t</i> is <i>d 1</i> over <i>v 1</i>. Then when we write the distance that the player travels, which is <i>d 2</i>, it will be <i>v 2</i> times <i>t</i>—the same time. We can substitute <i>d 1</i> over <i>v 1</i> in place of <i>t</i>. So I have 0.075 meters per second— velocity of the player—multiplied by 15.0 meters—distance covered for the puck— divided by the puck's velocity of 45.0 giving 2.5 centimeters is the distance covered by the player in the time it takes for the puck to reach the net.