1 dimensional elastic collision

AI Thread Summary
In a perfectly elastic collision between a 0.415-kg hockey puck moving at 2.65 m/s and a 0.910-kg puck at rest, the resulting velocities are -0.990 m/s for the lighter puck and 1.66 m/s for the heavier puck. There is a discussion about whether the lighter puck should have a higher velocity magnitude after the collision, given the conservation of kinetic energy. It is clarified that the distribution of kinetic energy depends on the mass ratio of the two pucks. The conversation emphasizes that the heavier puck can indeed move faster under certain conditions. Ultimately, the analysis confirms that the outcomes align with the principles of elastic collisions.
IniquiTrance
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A 0.415-kg hockey puck, moving east with a speed of 2.65 m/s, has a head-on collision with a 0.910-kg puck initially at rest.
Assuming a perfectly elastic collision, what will be the velocity of each object after the collision?
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The solution is:

v ' (lighter puck) , v' (heavier puck) = -0.990, 1.66 m/s, respectively.

My question is, wouldn't the lighter puck be expected to have a higher velocity magnitude than the heavier puck following the collision?
 
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For perfectly elastic collisions, kinetic energy is conserved.

K_i=K_1+K_2
 
Ok, i made that assumption when solving the problem.

I'm asking conceptually, wouldn't the heavier object be expected to move with a smaller velocity magnitude?
 
IniquiTrance said:
Ok, i made that assumption when solving the problem.

I'm asking conceptually, wouldn't the heavier object be expected to move with a smaller velocity magnitude?

Sorry, I didn't read your question properly :redface:

That will only be the case if the initial kinetic energy is divided equally between the two objects, i.e. when after the collision

\frac{1}{2}m_1v_1^2=\frac{1}{2}m_2v_2^2

so that

\frac{m_1}{m_2}=\frac{v_2^2}{v_1^2}
 
Ah gotcha. Thanks for the help! :smile:
 
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