# Homework Help: Collision Question: ( ) Exam tomorrow

1. Jul 28, 2013

### Dreaam

Collision Question: (URGENT) Exam tomorrow

Hi all,
I have an exam on collision and projectile motion tomorrow, and was wondering if anyone could answer this question: I get an answer, but my friend gets a completely different answer. I was wondering if anyone could just clarify and show us the correct answer. All help greatly appreciated. Thank you for your time.

1. The problem statement, all variables and given/known data

A black billiard ball travels up the centre line at 1.5 m/s. It's mass is 250g, it strikes a white ball of the same mass which is propelled at an angle of 35 degrees to the left of the original direction of the black ball. The white ball final speed is 0.9 m/s. What is the final velocity of the black ball?

2. Relevant equations

m1u1 + m2u2 = m1v1 + m2v2
tanθ = sinθ/cosθ

3. The attempt at a solution

This is what I did:

Find x and y components of the unknown final velocity's vector's magnitude.

Let Vb = Velocity of black ball
θb = angle the black ball is moved in after collision

x-component:
m1u1 + m2u2 = m1v1 + m2v2
(0.25 * 0) + (0.25 * 1.5 cos 90) = (0.25 * 0.9 cos 125) + (0.25 Vb cos θb)
0 = -0.13 + 0.25 Vb cos θb
0.13 = 0.25 Vb cos θb
Vb cos θb = 0.13 / 0.25 = 0.52 m/s

y-component
m1u1 + m2u2 = m1v1 + m2v2
(0.25 * 0) + (0.25 * 1.5 sin 90) = (0.25 * 0.9 sin 125) + (0.25 * Vb sin θb)
0 + 0.375 = 0.18 + 0.25 Vb sin θb
0.375 - 0.18 = 0.25 Vb sin θb
0.195 = 0.25 Vb sin θb
Vb sin θb = 0.195/ 0.25 = 0.78 m/s

So from this, using the magnitudes a right angle triangle can be formed:
positive 0.78 (north direction) being the vertical side
positive 0.52 (east direction) being the horizontal side

to find the resultant vector
r = sqroot (0.52^2 + 0.78^2)
r = 0.87 m/s

to find the direction:
tanθ = sin θ / cosθ
tanθb = vb sin θb / vb cos θb
tanθb = 0.78 / 0.52
θb = tan^-1 (0.78 / 0.52)
θb = 56.30

Therefore: the black ball after collision is moving at 0.87 m/s in a direction of 56.30 degrees north of east.

Last edited: Jul 28, 2013
2. Jul 28, 2013

### siddharth23

Could you please post a picture showing the directions and θ?

3. Jul 28, 2013

### Dreaam

I've attached the file.

Edit: Sorry the angle should be 35.

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4. Jul 28, 2013

### haruspex

You shouldn't still have a .25 in there.
Here's a really good piece of advice: work entirely symbolically until the last step. Only then plug in the actual numbers. This makes it much easier to find mistakes and for other people to follow what you are doing. In the above, you might then have spotted that the dimensions didn't make sense, having removed mass from all except one term. It also helps to preserve precision and minimise calculation because you often get cancellation in the algebra.

5. Jul 28, 2013

### Dreaam

Thank you for your feedback haruspex, my question is: even if I included the 0.25 values, wouldn't the outcome still be the same? Or have I made a mistake along the way?

6. Jul 28, 2013

### siddharth23

You've jumbled up sin and cos mate. x-component will have sinθ and y-component will have cosθ. Change that and solve again.

7. Jul 28, 2013

### haruspex

No, it is an error that will change the answer. You cancelled the 0.25 out of three terms but missed the fourth. You made the same mistake in the y direction.

siddharth23 is also right - you've mixed up sin and cos on the RHS of your equations.

8. Jul 28, 2013

### Dreaam

I see where I went wrong. Thank you everyone for your feedback.

9. Jul 28, 2013

### siddharth23

Anytime mate..