Qxi.xli
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Two identical particles of mass m are projected towards each other along the same straight line on a smooth horizontal surface with speeds 2u and 3u. After the collision the directions of motion of both particles are reversed. Show that this implies that the coefficient of restitution e satisfies the inequality e > 1/5

The answers are on page 7: https://activeteach-prod.resource.pe...b_fm1_ex4a.pdf
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dw it's not a virus lmao


I just don't understand why u must be>1 (4th line from the bottom lol) Please could someone explain this? They haven't specified that u>1 anywhere in the question either.

tyy xx
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tej3141
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I think that's an error and its mean to say u>0. That's the only logical reason I can see. But I havent started further mech yet so I may be wrong
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RDKGames
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(Original post by Qxi.xli)
Two identical particles of mass m are projected towards each other along the same straight line on a smooth horizontal surface with speeds 2u and 3u. After the collision the directions of motion of both particles are reversed. Show that this implies that the coefficient of restitution e satisfies the inequality e > 1/5

The answers are on page 7: https://activeteach-prod.resource.pe...b_fm1_ex4a.pdf
Spoiler:
Show
dw it's not a virus lmao


I just don't understand why u must be>1 (4th line from the bottom lol) Please could someone explain this? They haven't specified that u>1 anywhere in the question either.

tyy xx
Probably a typo and should say u>0
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Qxi.xli
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Also, I just had a quick question about Newton's law of restitution, if two objects are moving towards each other with say a speed of 2 and 5, why is the speed of approach of particles in this case 7? Wouldn't the overall/relative speed be 5-3 so 2?
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Qxi.xli
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(Original post by tej3141)
I think that's an error and its mean to say u>0. That's the only logical reason I can see. But I havent started further mech yet so I may be wrong
(Original post by RDKGames)
Probably a typo and should say u>0
thank youu xx
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tej3141
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(Original post by Qxi.xli)
Also, I just had a quick question about Newton's law of restitution, if two objects are moving towards each other with say a speed of 2 and 5, why is the speed of approach of particles in this case 7? Wouldn't the overall/relative speed be 5-3 so 2?
It's kind of intuitive I think. Say 2 particles are moving together or away from each other. The relative speeds will be greater and so you add. But if you look at the example where the particles move in the same direction, if one moves at 4m/s and one moves at 1.5m/s you would minus to get the speed of approach as 2.5m/s. I hope that makes some sense
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Qxi.xli
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(Original post by tej3141)
It's kind of intuitive I think. Say 2 particles are moving together or away from each other. The relative speeds will be greater and so you add. But if you look at the example where the particles move in the same direction, if one moves at 4m/s and one moves at 1.5m/s you would minus to get the speed of approach as 2.5m/s. I hope that makes some sense
ah thanks, the last bit makes sense. But if they're moving together wouldn't you take the speeds away to find the speed of approach, if that makes sense? x
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tej3141
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(Original post by Qxi.xli)
ah thanks, the last bit makes sense. But if they're moving together wouldn't you take the speeds away to find the speed of approach, if that makes sense? x
This may sound weird but I like to think of 2 particles that move together as moving away.
But an easier alternative way to comprehend is that the velocities are opposite as the are moving in opposite directions (even though it is called speed of approach). For the particles with 2 and 5m/s moving together,on will have v=2 and one will have v=-5. So if you minus like you said you will get 2-(-5)=7m/s
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Qxi.xli
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(Original post by tej3141)
This may sound weird but I like to think of 2 particles that move together as moving away.
But an easier alternative way to comprehend is that the velocities are opposite as the are moving in opposite directions (even though it is called speed of approach). For the particles with 2 and 5m/s moving together,on will have v=2 and one will have v=-5. So if you minus like you said you will get 2-(-5)=7m/s
ohh makes sense thank you!!
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mqb2766
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(Original post by Qxi.xli)
ohh makes sense thank you!!
Another view on that, is to think about the relative motion of two objects, so consider an observer on one object. So if you have two trains approaching each other at 100mph, their relative speed is 200mph. When the trains pass, their relative speed is still 200mph even though they're moving away from each oher now. The relative direction of motion the train is the same, though the relative position is different depending on whether it is coming towards you (in front) or moving away (behnd you). However the speed (velocity) is not dependent on position.
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Qxi.xli
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(Original post by mqb2766)
Another view on that, is to think about the relative motion of two objects, so consider an observer on one object. So if you have two trains approaching each other at 100mph, their relative speed is 200mph. When the trains pass, their relative speed is still 200mph even though they're moving away from each oher now. The relative direction of motion the train is the same, though the relative position is different depending on whether it is coming towards you (in front) or moving away (behnd you). However the speed (velocity) is not dependent on position.
Ooh thank youu xx
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