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Elastic Collision

Spongebob*No*Pants

When an object of mass M and velocity u collides head-on elastically with a stationary object of mass m, show that mass m, moves off with velocity V2 given by:

V2 = (2M/M+m) u

I done:

Mu + m = MV1 + mV2

and making V2 subject of formula...

V2 = (Mu + m - MV1)/m

but from here got stuck.

any help will be appreciated.
+rep offered to most thorough answer.

Thanks!

Elastic collision means energy is conserved
so

\frac{1}{2}Mu^2 = \frac{1}{2}Mv_1^2 +\frac{1}{2}mv_2^2

momentum consideration gives:

Mu = Mv_1 + mv_2

Rearrange the momentum equation to get v1 in terms of u and v2. then use the energy equation to eliminate v1 from the system.

It's been a while since I did collisions but.... since its elastic then e=1.

Mu=MV1+mV2

V1=(Mu-mV2)/M

e = Seperation /apprach

So 1 = (V2 - ((Mu-mV2)/M) /u

u = (MV2-Mu+mV2)M

Mu = MV2-Mu+mV2

MV2+mV2 = 2Mu

V2 = (2Mu/M+m)

And you just factorise and take the u out.

In your first equation you did Mu+m....the seconf object is stationary so its momentum is 0, its just Mu not Mu +m...

So before the collision, energy and momentum of the system is:

E = \frac{1}{2} M u^2

P = Mu

After the collision, the mass M now has velocity v_1 and the mass it hit, m, has velocity v_2:

E = \frac{1}{2} M v_1^2 + \frac{1}{2} m v_2^2

P = Mv_1 + mv_2

Equating this two equations together before and after the collision since the collision is elastic:

Mu = Mv_1 + mv_2 \to v_1 = \frac{Mu-mv_2}{M}

Mu^2 = Mv_1^2 + mv_2^2 \to Mu^2 = M\left[\frac{Mu-mv_2}{M}\right]^2 + mv_2^2

---- (1)

expanding out the bracket:

M\left[\frac{Mu-mv_2}{M}\right]^2 = Mu^2 - 2muv_2 + \frac{m^2 v_2^2}{M}

So eq (1) becomes:

Mu^2 = Mu^2 - 2muv_2 +\frac{m^2 v_2^2}{M} +mv_2^2 \to v_2 = \frac{2Mu}{M+m}

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