# A-Level Physics help

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Thread starter 7 years ago
#1
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?
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7 years ago
#2
Think about the conservation of momentum.
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Thread starter 7 years ago
#3
(Original post by anatomical frog)
Think about the conservation of momentum.
i've got

Vab = Ua + 0 (as b was at rest)
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7 years ago
#4
(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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Thread starter 7 years ago
#5
(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.
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Thread starter 7 years ago
#6
can someone explain please?
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7 years ago
#7
Well we know that:

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

Can we simplify that down?
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Thread starter 7 years ago
#8
(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
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7 years ago
#9
(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
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Thread starter 7 years ago
#10
(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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Thread starter 7 years ago
#11
(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?
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7 years ago
#12
(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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Thread starter 7 years ago
#13
(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.
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7 years ago
#14
Oh Ok, that's fair enough. As long as you understand it now.
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Thread starter 7 years ago
#15
(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?
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7 years ago
#16
(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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Thread starter 7 years ago
#17
(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
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7 years ago
#18
(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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Thread starter 7 years ago
#19
(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
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7 years ago
#20
(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.
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