- #1

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Ball A: mass=10 velocity=20

Ball B: mass=30 velocity=5

- Thread starter Physicsy
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- #1

- 8

- 0

Ball A: mass=10 velocity=20

Ball B: mass=30 velocity=5

- #2

- 8

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Ball A: 5

Ball B: 10

Could someone please confirm this for me.

- #3

- 648

- 2

It depends on how elastic the collision is. There are many possible solutions to the problem.

- #4

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The collision is perfectly elastic and all momentum is conversed

- #5

- 648

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Before they collide they are approaching with a relative velocity of 20-5 = 15m/s

This means that, in addition to the momentum being conserved, you also know that the balls will separate such that relative velocity is still 15m/s

So, for example, if ball B moves with velocity 10 m/s after the collision, then ball A would need to move at -5 m/s after the collision. That is, at 5m/s in the opposite direction to its original direction.

How have you made this calculation?

The answers cannot be correct if the collision is perfectly elastic.

- #6

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I then remeabered that the momentum behind both balls together must always be the same. So i found the momentum of both balls, Ball A=200 Ball B=150, then added them together, 200+150=350. I then took away the after collision momentum of Ball B,

10*30=300 350-300=50, and then figured out the after collision velocity of Ball A through this, 50/10=5.

Unless i made some sort of mistake or everything above is pure bull, i thought the answer would have been,

Ball A=5 (I understand that this is really -5 as the ball would reverse in direction)

Ball B=10

Also when i say perfectly elastic i mean that no kinetic energy during the collision is converted into heat between molecules of the balls, so all energy remains in the momentum.

- #7

- 648

- 2

If you check your values and work out the total kinetic before and after the collision, you will see that it is not conserved for the vlaues you have calculated.

[I worked it out that there is 2375J before the collision, and 1625J after, if your values are used.]

The way to do this is to 1st apply conservation of momentum

MaUa + MbUb = MaVa + MbVb

Ma=10kg, Mb=30kg, Ua=20m/s, Ub=5m/s, Va and Vb are the velocities after the collision.

This gives

20 x 10 + 30 x 5 = MaVa + MbVb

350 = 10Va + 30Vb

Then write down the relative velocity equation

Ua-Ub = -(Va - Vb)

15 = Vb-Va

From this get an expression for either Va or Vb and substitute it into the momentum equation to get one of the velocities.

Then it's easy to find the other velocity.

- #8

- 99

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I think you may use law of conservation of momentum and law of conservation of KE.

Ball A: mass=10 velocity=20

Ball B: mass=30 velocity=5

- #9

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Thanks guys, I understand why it can't be right now.

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