Calculating Momentum and Energy Conservation in Elastic Collisions

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To solve for momentum and energy conservation in elastic collisions, apply conservation principles in both x and y directions, as momentum is a vector quantity with directional components. The total momentum before and after the collision must remain equal in each direction. Additionally, kinetic energy must be conserved, meaning the total kinetic energy before the collision must equal the total after. Establish equations for momentum in both directions and for energy conservation, involving the final velocities and angles of the masses. Starting in the center of momentum frame simplifies calculations, as the masses will have equal-but-opposite momentum before and after the collision.
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Homework Statement
Can you find the magnitude of V1' by applying conservation of linear momentum along the direction of motion of m1 after collision ?
Relevant Equations
(M1)U1 +(M2)U2 = (M1)V1 +(M2)V2
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I know how to solve along x and y-axis but i can't think of how to start solving in the dricection on m1.
 
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You could apply conservation of momentum in any direction. But using x and y has its advantages -- the equations are a bit simpler.
 
What do you mean by soliving in the direction of m1?
 
Momentum is a vector quantity. That means, it has components as does velocity. So you need to total up the momentum in each direction before, then after. It has to be the same before as after in each direction.

At the same time, assuming an elastic collision, you need to conserve energy. So before you have kinetic energy of m1. After you have a different kinetic energy for m1, and a new one for m2. The before and after need to be equal.

So you will get an x-direction momentum equation, a y-direction momentum equation, and an energy conservation equation. These will involve the magnitude of the after-velocity of each mass (two unknowns) and the two angles.

Here's a hint. Start in the center of momentum frame. In that frame the two masses start with equal-but-opposite momentum. And they finish with equal-but-opposite momentum at some other angle. Then get your final answer and transform back to the lab frame.
 
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