Calculating Momentum and Direction in a Skating Collision

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SUMMARY

This discussion focuses on calculating momentum and direction in a collision scenario involving two skaters and the recoil of a gun. The skater on the left has a mass of 80 kg and moves at 2 m/s, while the skater on the right has a mass of 60 kg and moves at 1.5 m/s. The total momentum before the collision is calculated as 250 kg m/s, but direction must be considered, leading to the use of negative values for one skater's velocity. The discussion also clarifies that momentum is a vector quantity, necessitating the consideration of direction in calculations, as demonstrated in the gun recoil example.

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99jolegg
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First off, this isn't homework, I need to re-learn GCSE physics from a textbook and am having trouble with a few aspects of momentum.

Homework Statement


Two people are skating towards each other. The person on the left has a mass of 80kg and is skating at 2 m/s. The person on the right is skating towards the person on the left at 1.5 m/s and has a mass of 60kg

At what velocity do they collide and in what direction do they move off in?

Homework Equations


Momentum = Mass x Velocity


The Attempt at a Solution


I multiplied the first skater's mass by their speed and then multiplied the second skater's mass by their speed and added them together to get a combined momentum.

Skater One: 80kg x 2m/s = 160 kg m/s.
Skater Two: 60kg x 1.5m/s = 90 kg m/s.

Total combined momentum of 250 kg m/s

The revision guide suggests:

1) Choose which direction is positive? What does this mean?
2) For some reason, they have worked it out as (80kg x 2m/s) + (60kg x -1.5m/s). Why is it minus? I don't understand why one of the speeds is then negative?

Thanks in advance.
 
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Momentum (and velocity) is a vector. Direction--and thus sign--counts! Choose one direction ("to the right", say) to be positive; make the other negative.
 
Doc Al said:
Momentum (and velocity) is a vector. Direction--and thus sign--counts! Choose one direction ("to the right", say) to be positive; make the other negative.

Thanks for the response. I understand that they are both vector quantities but I don't really understand the effect of making one momentum negative has on that.

Also, another question is about the recoil of a gun being fired, yet there are no negative figures used when working out the momentum. Why is this?

Cheers
 
99jolegg said:
Thanks for the response. I understand that they are both vector quantities but I don't really understand the effect of making one momentum negative has on that.
Since momentum is a vector, you can't add two momentums together (to find the total momentum of a system) without taking direction into account.

If a car moves to the right at 50 mph and a second identical car moves to the left with the same speed are their momentums the same? No. They have the same magnitude but not direction. What's the total momentum of both cars? Zero.
Also, another question is about the recoil of a gun being fired, yet there are no negative figures used when working out the momentum. Why is this?
Beats me. Provide the complete problem.
 
Thanks, kinda makes sense.

"A gun fires a bullet as shown. At what speed does the gun move backwards?"

Bullet Velocity = 150 m/s
Bullet Mass = 0.01 kg

Gun Velocity = V
Gun Mass = 1 kg

The book says you should work it out as below:

1) Choose which direction is positive, he chooses right as positive.
2) Total Momentum Before firing:
= 0kg
3) Total Momentum After firing:
= (0.01 kg x 150 m/s) + (1 x v)
Therefore: 1.5 kg m/s + v = 0
4) v = -1.5 m/s
5) As we established right is positive, and our answer is negative, the gun moves backwards (left) at 1.5 m/s.

Yet, none of the numbers are negative. Is that because the velocity of the gun is unknown at the beginning?
 
Not exactly sure of your question. The gun and bullet must move in opposite directions, so one of the velocities must be negative. Note that "speed" is just the magnitude of the velocity.
 
I understand - you don't need to make one of the numbers negative because the number you are looking for, velocity, is the answer, i.e. different to the first question.

I'll try some practice questions.

Thanks for your help.
 

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