Conservation Laws and Velocity Reversal in 1D Collisions

[skoczek77]

Homework Statement

A body of mass m moving with speed v hits a resting body of mass M. After an ideally elastic collision, the masses move in opposite directions with equal velocities. Give the ratio of the masses of bodies m/M (as a number). We neglect friction.

Relevant Equations

principle of conservation of momentum and kinetic energy

i dont know how

 

[Ibix]

What is the momentum and energy before the collision? What is the momentum and enetgy after the collision?

 

[skoczek77]

p=mv1 before and
p=-mv2 +Mv2 after

 

[PeroK]

p=mv1 before and
p=-mv2 +Mv2 after

That gives you an equation. Can you find another equation knowing that the collision is elastic?

 

[skoczek77]

with kinetic energy but idk how

 

[PeroK]

The homework help guidelines require you to make an effort and not just say idk.

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

Check your notes for kinetic energy and elastic collisions.

 

[Ibix]

p=mv1 before and
p=-mv2 +Mv2 after

Right. So what's the kinetic energy of a moving body? And hence, what are the equivalent conservation equations for energy?

 

[skoczek77]

I've already done it, tell me if it's good:

0,5·m·v1²=0,5·(M+m)·v2²

m·v1²=(M+m)·v2²


v2=(m·v1)/(M-m)

m·v1²=(M+m)·(m·v1)²/(M-m)²

m·v1²=(M+m)·m²·v1²/(M-m)²

1=(M+m)·m/(M-m)²

(M-m)²=m·M+m²

M²-2·M·m+m²=m·M+m²

M²-2·M·m=m·M

M-2·m=m

M=3·m

so m/M=1/3

 

[PeroK]

Yes, that's correct. I guess you never know when a sudden burst of algebraic creativity will strike!

 

[skoczek77]

yes, you are right, 2 hours ago I thought there was not enough data to solve it;
thank you very much for help
have a nice day ;)

 

[haruspex]

For future reference, in a fully elastic 1D collision, a very simple relationship can be deduced from the conservation laws: the velocity difference is reversed.
That is, if the initial velocities are $v_1, v_2$ and the final velocities $v'_1, v'_2$ then $v_1- v_2=v'_2- v'_1$.
In the present case, you have $v_2=0, v'_2=-v'_1$, so $v'_1=-\frac 12v_1$.
Combining that with momentum conservation gives the answer without involving quadratics.

For the imperfectly elastic version, see https://en.wikipedia.org/wiki/Coefficient_of_restitution

Btw, the question statement is wrong. Moving "in opposite directions with equal velocities" is not possible; equal speeds, yes.