# Video: Using the Conservation of Linear Momentum in Analyzing a Collision

A 90.0-kg 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?

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### Video Transcript

A 90.0-kilogram hockey player hits a 0.150-kilogram puck, giving the puck a velocity of 45.0 metres per second. 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 metres away?

As we approach this problem, the fact that weβre treating the ice as frictionless means we assume no energy loss due to motion. Letβs call the distance that the player recoils π; thatβs what weβre trying to solve for. The critical information weβre given in this problem statement includes the mass of the hockey player, 90.0 kilograms, the mass of the puck, 0.150 kilograms, the final velocity of the puck, 45.0 metres per second. And weβre told that both the player and the puck are initially at rest. π, the distance the player recoils, takes place over the time that it takes the puck to travel 15.0 metres.

To solve this problem, weβll rely on the conservation of momentum. So letβs recall that relationship now. The law of conservation of momentum tells us that the initial momentum in a system, π sub π, is equal to the final momentum in that system, π sub π. We can expand out this equation by writing πππ£π equals πππ£π.

Letβs apply this relationship to our scenario. We have a system consisting of two objects: the hockey player and the puck. Our initial condition is before the hockey player has hit the puck and our final condition is after the puck has been hit.

So letβs write out our full conservation of momentum equation including both the player and the puck. Weβve abbreviated the mass of the player π sub π, the mass of the puck π sub π’, the initial velocity of the player π£ sub ππ, the initial velocity of the puck π£ sub ππ’. And weβre also given the final velocity of the puck, π£ sub ππ’.

When we write these terms into our momentum equation, we find this equation which has four terms. Looking at the terms on the left side of our equation, we see they include π£ sub ππ and π£ sub ππ’, the initial speed of the puck and player. Both of those are given as zero, which means that both terms on the left side of our equation are zero. In other words, the initial momentum of our system is zero.

When we look at the right side of our equation, we see that these terms involve four values: the mass of the player, the final speed of the player, the mass of the puck, and the final speed of the puck. The only value of those four that we donβt know already is π£ sub ππ, the final speed of the player.

Letβs rearrange this equation to solve for that speed. First, we subtract π sub π’, π£ sub ππ’ from both sides, which cancels out that term on the right-hand side of our equation. We then divide both sides of our equation by π sub π, the mass of the player, which cancels out that term on the right side of our equation, leaving π£ sub ππ. This leaves us with an equation which says π£ sub ππ, the final speed of the player, is equal to negative the mass of the puck times the final speed of the puck divided by the mass of the player.

When we plug these values in to our equation, we find a final speed of the player of 0.075 metres per second. Weβve solved for the final velocity of the player according to the conservation of momentum, but we were asked to solve for the distance that the player recoils, as a result of the interaction with the puck. Recall that speed in general is defined as the distance travelled divided by the time it took to travel that distance.

To solve for the distance the player recoils, weβll need to know the time over which that recoil happens. To solve for that, letβs recall that weβre told that the puck travels 15.0 metres in order to reach the goal. Weβll call that π sub π’. If the puck is moving with a final speed of 45.0 metres per second and travels a distance of 15.0 metres, then using our relationship between speed, distance, and time, we see that the time the puck took to travel this distance π‘ sub π’ is equal to the distance travelled π sub π’ divided by the speed of the puck π£ sub ππ’.

When we plug in the values given for these terms and perform this division, we find that π‘ sub π’, the time it takes the puck to reach the net after being hit by the hockey player, is 0.333 seconds. That time π‘ sub π’ is also the time over which the player recoils. We can use that time in a new speed equals distance over time equation. To solve for π, the distance the hockey player recoils, the relationship between the three terms is π equals π£ sub ππ multiplied by π‘ sub π’.

We know both of those values and when we plug them in and multiply them together, we find a displacement of negative 2.50 centimetres, which means that the magnitude of the distance weβve recoiled is 2.50 centimetres. Thatβs the distance that the hockey player recoils back over the time that the puck travels 15.0 metres.