# Momentum problem

1. Jul 17, 2006

### Hollysmoke

Two balls A and B having different but unkown masses collide. A is initially at rest and B has a speed v moving North. After collision B has a speed v/2 and moves perpendicular (East) to the original motion. Find the direction of motion of A after collision. Is it possible to determine the speed of A from the information given? Explain.

What I did was I drew out a grid with x and y axis, and I figured that the y-axis would have to cancel out while momentum is conserved in the x-axis, so the motion is going to be somewhere in hte lower left quadrant. But after that, I'm stuck.

2. Jul 17, 2006

### Andrew Mason

From conservation of momentum:

$$m_a\vec{v_{af}} + m_b\vec{v_{bf}} = m_b\vec{v_{bi}}$$

So:

(1) $$m_av_{af}\cos\theta = -m_bv/2$$

(2) $$m_av_{af}\sin\theta = m_bv$$

Dividing 2 by 1:

$$\tan\theta = -2$$

So you know the angle.

If you also assume that there is conservation of energy, you can determine the speed of a in terms of the relative masses:

$$\frac{1}{2}m_bv^2 = \frac{1}{2}m_bv^2/4 + \frac{1}{2}m_av_a^2$$

$$m_av_a^2 = \frac{3}{4}m_bv^2$$

which gives us:

$$v_a = \sqrt{\frac{3m_b}{4m_a}}v$$

But I think that is as far as you can go unless you know the relative masses.

AM

Edit: tan = -2

Last edited: Jul 18, 2006
3. Jul 17, 2006

### Office_Shredder

Staff Emeritus
You've got it backwards there. Assuming north is positive y-axis, momentum must be conserved vertically, and horizontally it must cancel out. So the new momentum must be in the upper left quadrant.

Does that help at all?