# A-Level Physics help

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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?

calculate Vab

why is it Ua/2?

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(Original post by

Think about the conservation of momentum.

**anatomical frog**)Think about the conservation of momentum.

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

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(Original post by

Momentum is mass x velocity so you need mass in there too.

**Stonebridge**)Momentum is mass x velocity so you need mass in there too.

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(Original post by

Well we know that:

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

Can we simplify that down?

**anatomical frog**)Well we know that:

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

Can we simplify that down?

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

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#9

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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(Original post by

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

**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

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**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

mUa/2m = Vab

if you cancel the m

Ua/m = Vab?

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#12

(Original post by

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?

**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?

Ua = 2Vab

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

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(Original post by

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.

**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.

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(Original post by

Oh Ok, that's fair enough. As long as you understand it now.

**anatomical frog**)Oh Ok, that's fair enough. As long as you understand it now.

what fraction of kinetic energy is lost?

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#16

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

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(Original post by

Well, do you know the formula for kinetic energy?

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

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

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#18

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

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]

**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]

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

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#20

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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