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OCR Momentum

A block of mass 3kg travelling at 5ms-1 catches up another block of mass 7kg travelling at 2 ms-1 along the same line and the same direction.

The difference between the speeds of the two blocks after the collision cannot be greater than the difference of the speeds before the collision. Deduce that the speed of the first block after the collision must satisfy the inequalities 0.8<=v1<=2.9

I had v2=(29-3v1)/5, and v1-3<=v2<=v1+3, which eventually leads me to 14<=8v<=44. This clearly isn't right, but I really don't know what to do. Help appreciated.
Original post by Julii92
A block of mass 3kg travelling at 5ms-1 catches up another block of mass 7kg travelling at 2 ms-1 along the same line and the same direction.

The difference between the speeds of the two blocks after the collision cannot be greater than the difference of the speeds before the collision. Deduce that the speed of the first block after the collision must satisfy the inequalities 0.8<=v1<=2.9

I had v2=(29-3v1)/5, and v1-3<=v2<=v1+3, which eventually leads me to 14<=8v<=44. This clearly isn't right, but I really don't know what to do. Help appreciated.


Define a number e such that e=speedofseparationspeedofapproache= \frac{speed of separation}{speed of approach}. Work from there
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Julii92
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Original post by TheMagicMan
Define a number e such that e=speedofseparationspeedofapproache= \frac{speed of separation}{speed of approach}. Work from there


Thnanks, but I'm still not sure how to proceed from there. Can you tell me any more?
Original post by Julii92

I had v2=(29-3v1)/5


Why do you have it over 5?

You also need to know that the speed of separation must be >=0.
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Julii92
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Original post by ghostwalker
Why do you have it over 5?

You also need to know that the speed of separation must be >=0.


m1u1 + m2u2=m1v1 + m2v2
3x5 + 7x2 = 3v1 + 5v2
29=3v1+5v2
29-3v1=5v2
v2=(29-3v1)/5

How does the speed of separation being >=0 help?
Original post by Julii92
m1u1 + m2u2=m1v1 + m2v2
3x5 + 7x2 = 3v1 + 5v2
29=3v1+5v2
29-3v1=5v2
v2=(29-3v1)/5


m2=7m_2=7


How does the speed of separation being >=0 help?


The inequality you have, although it gives two limits (if I recall), only one of them matches the restriction you're aiming for.

You need the additional bit to get the second limit.
(edited 11 years ago)
Reply 6
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Julii92
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Original post by ghostwalker
m2=7m_2=7


Ah. Now I feel silly.


Original post by ghostwalker
The inequality you have, although it gives two limits (if I recall), only one of them matches the restriction you're aiming for.

You need the additional bit to get the second limit.


Okay, I'll have a go at using this.

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