Finding loss of energy in collision

[Saitama]

Homework Statement

A particle of mass $m_1$ experienced a perfectly elastic collision with a stationary particle of mass $m_2$. What fraction of the kinetic energy does the striking particle lose, if it recoils at right angles to its original motion direction.

(Ans: $2m_1/(m_1+m_2)$ )

Homework Equations

The Attempt at a Solution

Let the initial velocity of $m_1$ be $v$ and let the x-axis be along the initial direction of motion. After collision, the first particle flies off at right angles and let that direction be y-axis. The vertical component of velocity of $m_2$ after collision has the direction opposite to that of $m_1$. Conserving momentum in x direction:
$$m_1v=m_2v_{2x}$$
Conserving momentum in y direction:
$$m_1v_1=m_2v_{2y}$$
where $v_1$ is the final velocity of $m_1$. I still need one more equation.

Any help is appreciated. Thanks!

 

[Curious3141]

What quantity, apart from momentum, is conserved in a perfectly elastic collision?

 

[Saitama]

What quantity, apart from momentum, is conserved in a perfectly elastic collision?

Energy. But that gives a very dirty equation. I have seen some problems on one dimensional collisions where coefficient of restitution is used to find another equation. Is it possible to apply the same here as I think it reduces the algebra work greatly.

 

[Curious3141]

Energy. But that gives a very dirty equation. I have seen some problems on one dimensional collisions where coefficient of restitution is used to find another equation. Is it possible to apply the same here as I think it reduces the algebra work greatly.

Energy, as in total energy, is *always* conserved.

In a perfectly elastic collision, kinetic energy is specifically conserved.

The algebra is fairly easy to work out here. Took me less than 10 lines and barely 5 minutes.

Remember that the final speed of $m_2$ is given by the Pythagorean theorem. Deal only in squares of the velocity components, and everything simplifies quickly.

And always keep in mind what you're trying to find, which is the ratio $\displaystyle \frac{v^2 - v_1^2}{v^2}$.

 

[Saitama]

Energy, as in total energy, is *always* conserved.

In a perfectly elastic collision, kinetic energy is specifically conserved.

The algebra is fairly easy to work out here. Took me less than 10 lines and barely 5 minutes.

Remember that the final speed of $m_2$ is given by the Pythagorean theorem. Deal only in squares of the velocity components, and everything simplifies quickly.

And always keep in mind what you're trying to find, which is the ratio $\displaystyle \frac{v^2 - v_1^2}{v^2}$.

I have solved the problem using the energy approach, thanks a lot Curious!

 

[Curious3141]

I have solved the problem using the energy approach, thanks a lot Curious!

You're welcome, and I'm glad you solved it. Sorry I couldn't stay up to continue to help, but I need early nights as I've not been in the best of health lately.

 

[Saitama]

You're welcome, and I'm glad you solved it. Sorry I couldn't stay up to continue to help, but I need early nights as I've not been in the best of health lately.

I hope you get well soon. :)