elastic collision in two dimensions

• Dec 3rd 2007, 05:27 PM
Linnus
elastic collision in two dimensions
On a frictionless surface, a 0.35 kg puck moves horizontally to the right (at an angle of 0°) and a speed of 2.3 m/s. It collides with a 0.23 kg puck that is stationary. After the collision, the puck that was initially moving has a speed of 2.0 m/s and is moving at an angle of −32°. What is the velocity of the other puck after the collision?
• Dec 3rd 2007, 05:55 PM
topsquark
Quote:

Originally Posted by Linnus
On a frictionless surface, a 0.35 kg puck moves horizontally to the right (at an angle of 0°) and a speed of 2.3 m/s. It collides with a 0.23 kg puck that is stationary. After the collision, the puck that was initially moving has a speed of 2.0 m/s and is moving at an angle of −32°. What is the velocity of the other puck after the collision?

Nothing in the problem states that the collision is elastic. Thus I will not assume that it is. So no energy conservation. (Why did you write that it is elastic in the title?)

During the collision there are no net external forces on the two pucks. Thus we know that the momentum of the two pucks is conserved. Since this is a two dimensional problem, we can also note that the momentum of the system is conserved in both the x and y directions:
$m_1v_{10x} + m_2v_{20x} = m_1v_{1x} + m_2v_{2x}$
and
$m_1v_{10y} + m_2v_{20y} = m_1v_{1y} + m_2v_{2y}$

So let's define us some positive directions. (You could also define an origin, a good practice, but it will not be needed here.) I'm going to let there be a +x axis in the direction the 0.35 kg puck is moving in before the collision. (To the right, in other words.) I am going to define a +y direction "straight up." That is to say straight up when you are sketching this problem.

So calling puck 1 the 0.35 kg puck, and puck 2 the 0.23 kg puck, we know that
$v_{10x} = 2.3~m/s$
$v_{10y} = v_{20x} = v_{20y} = 0~m/s$

So
$m_1v_{10x} = m_1v_{1x} + m_2v_{2x}$
and
$0 = m_1v_{1y} + m_2v_{2y}$

We also know
$v_{1x} = 2.0 \cdot cos(32^o)$
$v_{1y} = -2.0 \cdot sin(32^o)$

So
$m_1v_{10x} = 2m_1~cos(32^o) + m_2v_{2x}$
and
$0 = -2m_1~sin(32^o) + m_2v_{2y}$

The first equation says:
$v_{2x} = \frac{m_1v_{10x} - 2m_1~cos(32^o)}{m_2} = 0.918984~m/s$

and the second says:
$v_{2y} = \frac{2m_1~sin(32^o)}{m_2} = 1.6128~m/s$

We want the velocity of the second puck. So the magnitude of the velocity will be:
$v_2 = \sqrt{v_{2x}^2 + v_{2y}^2} = 1.85625~m/s$

The velocity vector is in the first quadrant, so the angle the velocity makes with the +x axis is
$\theta = tan^{-1} \left ( \frac{v_{2y}}{v_{2x}} \right ) = 60.3252^o$

Now a little unfinished business. I'll leave it to you to calculate whether or not this collision was elastic. (Big hint: It wasn't.)

-Dan
• Dec 3rd 2007, 06:07 PM
Linnus
oh, that was the title that the problem was under ^^;;