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A-Level Physics help

trolley a moves with initial speed Ua towards trolley b of equal mass which is at rest, the trolleys stick together and move off as one speed, Vab

calculate Vab

why is it Ua/2?
(edited 10 years ago)

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Think about the conservation of momentum.
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Original post by anatomical frog
Think about the conservation of momentum.


i've got

Vab = Ua + 0 (as b was at rest)
(edited 10 years ago)
Original post by SexyNerd
i've got

Vab = Ua + 0 (as b was at res)


Momentum is mass x velocity so you need mass in there too.
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Original post by Stonebridge
Momentum is mass x velocity so you need mass in there too.


ok, but the answer in the back is Ua/2, how? it makes sense as they have the same mass but i cant figure out why algebraically.
(edited 10 years ago)
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can someone explain please?
Well we know that:

mUa (+0mb) = (m+m)Vab

Can we simplify that down?
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Original post by anatomical frog
Well we know that:

mUa (+0mb) = (m+m)Vab

Can we simplify that down?


take m as a common factor?

m (Ua + 0b) = (m+m)Vab
Original post by SexyNerd
take m as a common factor?

m (Ua + 0b) = (m+m)Vab




You're thinking along the right lines, but if we do:

mUa = 2mVab

we can just divide both sides by m, and rearrange.

Edit: To clarify, I put the brackets in to make sure you understood how I got that equation, not for any mathematical reason
(edited 10 years ago)
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Original post by anatomical frog
You're thinking along the right lines, but if we do:

mUa = 2mVab

we can just divide both sides by m, and rearrange.

Edit: To clarify, I put the brackets in to make sure you understood how I got that equation, not for any mathematical reason


got you, them not adding in m led me to believe it wasnt needed in the equation. i must remember from now on to keep the equation in mind and work from there.
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Original post by anatomical frog
You're thinking along the right lines, but if we do:

mUa = 2mVab

we can just divide both sides by m, and rearrange.

Edit: To clarify, I put the brackets in to make sure you understood how I got that equation, not for any mathematical reason


because unless you include m, you cant get the right answer so why did the textbook leave it out?

mUa/2m = Vab

if you cancel the m

Ua/m = Vab?
Original post by SexyNerd
because unless you include m, you cant get the right answer so why did the textbook leave it out?

mUa/2m = Vab

if you cancel the m

Ua/m = Vab?


If you have mUa = 2mVab and you divide by m, you just get:

Ua = 2Vab
so Vab = Ua/2 which was the answer you were looking for.
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Original post by anatomical frog
If you have mUa = 2mVab and you divide by m, you just get:

Ua = 2Vab
so Vab = Ua/2 which was the answer you were looking for.


of course, pardon me, but like i said, they didn't mention m in the textbook, hence my confusion.
Oh Ok, that's fair enough. As long as you understand it now.
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Original post by anatomical frog
Oh Ok, that's fair enough. As long as you understand it now.


i do, thank you.

what fraction of kinetic energy is lost?
Original post by SexyNerd
i do, thank you.

what fraction of kinetic energy is lost?


Well, do you know the formula for kinetic energy?

If so, calculate the (end k.e - initial k.e) / the initial k.e
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Original post by anatomical frog
Well, do you know the formula for kinetic energy?

If so, calculate the (end k.e - initial k.e) / the initial k.e


i know the equation,

i get 1/2 MUa^2 = MVab^2
Original post by SexyNerd
i know the equation,

i get 1/2 MUa^2 = MVab^2



That doesn't work, as you shouldn't have the two things equaling.

What is the kinetic energy of trolley A before the collision?
What is the kinetic energy of trolley B before the collision?
What is the kinetic energy of AB after the collision?

Then you need to work it out so you get:

energy lost = (answer)

[I can see that you're thinking along the right lines though]
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Original post by anatomical frog
That doesn't work, as you shouldn't have the two things equaling.

What is the kinetic energy of trolley A before the collision?
What is the kinetic energy of trolley B before the collision?
What is the kinetic energy of AB after the collision?

Then you need to work it out so you get:

energy lost = (answer)

[I can see that you're thinking along the right lines though]


so

1/2 MUa^2 + 0 = 1/2 MVab^2 + 1/2 MVab^2
Original post by SexyNerd
so

1/2 MUa^2 + 0 = 1/2 MVab^2 + 1/2 MVab^2



That would only be correct if the collision was elastic (i.e: No kinetic energy was lost).
Also, you only have one moving object after the collision, so merge the two equations.

The equation is:

1/2 MUa^2 + 0 = MVab^2 + k.e lost


So find k.e lost. Then find that as a fraction of the total energy.
(edited 10 years ago)

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