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M2 Collisions

Hey, so heres the question I'm stuck on (part b)

A particle P of mass 3m is moving with speed 2u in a straight line on a smooth horizontal plane. The particle P collides directly with a particle Q of mass 4m moving on the plane with speed u in the opposite direction to p. The coefficient of restitution between p and Q is e

a.) Find the speed of Q immediately after the collision

I calculated this to be Vq=(1/7)(2+9e)u

Given that the direction of motion of P is reversed by the collision,

b) find the range of possible values of e

I need some help on this one. Thanks TSR
Reply 1
Original post by xyz9856
Hey, so heres the question I'm stuck on (part b)

A particle P of mass 3m is moving with speed 2u in a straight line on a smooth horizontal plane. The particle P collides directly with a particle Q of mass 4m moving on the plane with speed u in the opposite direction to p. The coefficient of restitution between p and Q is e

a.) Find the speed of Q immediately after the collision

I calculated this to be Vq=(1/7)(2+9e)u

Given that the direction of motion of P is reversed by the collision,

b) find the range of possible values of e

I need some help on this one. Thanks TSR


What's the velocity of P after the collision? (In terms of e) If it reverses direction shou that velocity be > 0 or < 0?
Reply 2
Original post by Zacken
What's the velocity of P after the collision? (In terms of e) If it reverses direction shou that velocity be > 0 or < 0?


since i set P as positive before it should be <0 after collision?
Reply 3
Original post by xyz9856
since i set P as positive before it should be <0 after collision?


Yeah, so find VpV_p in terms of ee, then set Vp<0V_p < 0 and solve for ee. For example, if I found that Vp=3e0.5V_p = 3e - 0.5 then I can say that 3e0.5<0e<0.533e - 0.5 < 0 \Rightarrow e < \frac{0.5}{3} (this is just an example, your VpV_p will be something different that you need to work out for yourself, do you know how to?)- remember to make sure that 0<e<10 <e < 1. Clear?
Reply 4
Original post by Zacken
Yeah, so find VpV_p in terms of ee, then set Vp<0V_p < 0 and solve for ee. For example, if I found that Vp=3e0.5V_p = 3e - 0.5 then I can say that 3e0.5<0e<0.533e - 0.5 < 0 \Rightarrow e < \frac{0.5}{3} (this is just an example, your VpV_p will be something different that you need to work out for yourself, do you know how to?)- remember to make sure that 0<e<10 <e < 1. Clear?


finally, that took me quite some time to get my head around. 1/6<e<1 was my answer. Im finding M2 and M1 quite tricky. How do I get myself to 100UMS standard between now and the exams (I believe they are in June, 8th and 17th). Thank you.
Reply 5
Original post by xyz9856
finally, that took me quite some time to get my head around. 1/6<e<1 was my answer. Im finding M2 and M1 quite tricky.


First class work.

How do I get myself to 100UMS standard between now and the exams (I believe they are in June, 8th and 17th). Thank you.


Practice.
Reply 6
Original post by Zacken
First class work.



Practice.



thanks Zacken, good luck with your exams also. I hope to be in a similar place to you at this time next year, except with engineering :smile:
Reply 7
Original post by xyz9856
thanks Zacken, good luck with your exams also. I hope to be in a similar place to you at this time next year, except with engineering :smile:


No problem - glad I helped. Good luck with your exams as well! TSR is here for any more questions you have. :biggrin:
That's very kind of you to say. :smile:
Original post by xyz9856
since i set P as positive before it should be <0 after collision?


How do you know if p is <0 or >0 after the collision?
Reply 9
Original post by mohamed-essa
How do you know if p is <0 or >0 after the collision?


Depends on whether you said p>0 before the collision. If you said p > 0 before then it's p < 0 after. If you said p < 0 before than it's p > 0 after.
Original post by Zacken
Depends on whether you said p>0 before the collision. If you said p > 0 before then it's p < 0 after. If you said p < 0 before than it's p > 0 after.


Thanks man.
Reply 11
Original post by mohamed-essa
Thanks man.


You're welcome!

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