Momentum Problem: Solving A and B's Post-Collision Direction and Speed

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The discussion focuses on the momentum problem involving two colliding balls, A and B, where A is initially at rest and B moves North with speed v. After the collision, B moves East with speed v/2. The conservation of momentum equations are established, leading to the conclusion that the angle of A's motion can be determined as tan(θ) = -2, indicating A moves in the upper left quadrant. However, the exact speed of A cannot be determined without knowing the relative masses of the balls.

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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.
 
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Hollysmoke said:
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.
From conservation of momentum:

[tex]m_a\vec{v_{af}} + m_b\vec{v_{bf}} = m_b\vec{v_{bi}}[/tex]

So:

(1) [tex]m_av_{af}\cos\theta = -m_bv/2[/tex]

(2) [tex]m_av_{af}\sin\theta = m_bv[/tex]

Dividing 2 by 1:

[tex]\tan\theta = -2[/tex]

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:

[tex]\frac{1}{2}m_bv^2 = \frac{1}{2}m_bv^2/4 + \frac{1}{2}m_av_a^2[/tex]

[tex]m_av_a^2 = \frac{3}{4}m_bv^2[/tex]

which gives us:

[tex]v_a = \sqrt{\frac{3m_b}{4m_a}}v[/tex]

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

AM

Edit: tan = -2
 
Last edited:
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.

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?
 

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