# Kinetic energy and momentum

Theres some calculations in this problem that I just dont get...

Start with the equations for initial and final momenta and kinetic energies and derive the

theoretical equation for the ratio fo $$K_f$$ to $$K_i$$

$$P_i=Mv_i$$
$$P_f=(M+m)v_f$$
$$K_i=1/2Mv_i^2$$
$$K_f=1/2(M+m)V_f^2$$
$$K_f/K_i=1/2(M+m)v_f^2/1/2Mv_i^2=M/(M+m)$$ this part I dont get. I only get

$$(M+m)/M$$ and im assuming that $$v_f$$ and $$v_i$$

cancel out...

I basically solved for $$M=P_i/v_i$$ and $$(M+m)=P_f/v_f$$
I plug it in to kinetic energy equations and get $$K_f/K_i=P_f/P_i$$ I guess $$v_f$$ and $$v_i$$ cancel out? Is this answer correct?

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I dont understand the intial question, are two bodies inelastically colliding? Could you be more specific as to the problem please.

Yes, it is an inelastic collision.

Also a seperate question is what would the difference between an elastic and inelastic collision?

the intial and final velocities are not going to be the same, in general. A simple example is a fly hitting a winsheild. Its traveling with its initial speed of lets say 2m/s. It hits the truck, and gets stuck on the winsheild, now its moving at the speed of the truck, maybe 30m/s.

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jtbell
Mentor
Meteo said:
difference between an elastic and inelastic collision?
In an elastic collision, the total kinetic energy is conserved. In an inelastic collision, the total kinetic energy is not conserved.

I think the most you can do is simplify the ratio as:

$$(1 + \frac{m}{M}) (\frac{V_f}{V_i})^2$$

but if you want it in terms of intial and final momentum it will be:

$$( \frac {M}{M+m})( \frac{ P_f}{P_i})^2$$

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In an elastic collision , the initial and kinetic energies remain the same But in inelastic collision , some of the initial KE is lost to the surroundings as other forms of energy like heat/sound , but in both the cases , the Total energy is always conserved.

BJ