Calculating Linear Momentum in Elastic Collision of Two Balls

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SUMMARY

The discussion focuses on calculating the linear momentum in an elastic collision between two balls, where one ball has a mass of 2m and is stationary, while the other has a mass of m and strikes it. The equations derived include the conservation of momentum: M1v1 + M2v2 = M1u1 + M2u2, and the conservation of kinetic energy: (1/2)(M1v1^2 + M2v2^2) = (1/2)(M1u1^2 + M2u2^2). The user concludes that the ratio of linear momentums can be determined, yielding a result of -1, assuming a head-on collision. Adjustments for non-head-on collisions require angle considerations.

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Two balls hit each other.

One ball with a mass of 2m standing still relative to the earth. The other ball has a mass of 1m and hits the ball of 2m at a certain speed. The collision is elastic.

Could you calculate the proportion of Linear Momentum between both balls after the collision with this information?
 
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M1 = 2m
M2 = m
v - speed before collision
u - speed after collision

[tex]M_1v_1 + M_2v_2 = M_1u_1 + M_2u_2[/tex]

Cancelling the mass and zero speed we get:

[tex]v_2 = 2u_1 + u_2[/tex]

Second equation:

[tex]\frac{1}{2}(M_1v_1^2 + M_2v_2^2) = \frac{1}{2}(M_1u_1^2 + M_2u_2^2)[/tex]

Cancelling the mass and zero speed we get:

[tex]v_2^2 = 2u_1^2 + u_2^2[/tex]

Two equations with two variables, you can find both u1 and u2 and therefore the ratio of linear momentums of both balls. I think it turns out to be -1...

(By the way, I assumed head-on collision here. If it's not you also need the angle.)
 
Last edited:
Chen said:
(By the way, I assumed head-on collision here. If it's not you also need the angle.)

Thanks!... it's actually not a head-on collision, but I figured out I could manage to fit in the angle myself. If I experience difficulties I'll let you know. :cool:
 

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