# Completely Inelastic collision

• ~angel~
In summary, we have discussed the collision of two particles with perpendicular motions, where one has mass m and momentum of magnitude 2p and the other has mass 2m and momentum of magnitude p. After the collision, the particles "trade" their momenta and move in the opposite direction. The kinetic energy lost in this collision is found to be 3p^2/4m. In a second scenario, where the particles collide completely inelastically, the kinetic energy lost is found to be 17p^2/12m.
~angel~

Two particles move perpendicular to each other until they collide. View Figure Particle 1 has mass m and momentum of magnitude 2p, and particle 2 has mass 2m and momentum of magnitude p. Note: Magnitudes are not drawn to scale in any of the figures.

a. Suppose that after the collision, the particles "trade" their momenta, as shown in the figure. That is, particle 1 now has magnitude of momentum p, and particle 2 has magnitude of momentum 2p; furthermore, each particle is now moving in the direction in which the other had been moving. How much kinetic energy, K_lost, is lost in the collision?

b.Consider an alternative situation: This time the particles collide completely inelastically. How much kinetic energy K_lost is lost in this case?

I've already found the answer to part a, which is 3p^2/4m, and as I was trying to work out the answer to b, I ended up with the asme answer, and I followed all the "instructions" in my textbook. Please help.

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In part (b), the particles collide "completely inelastically" which means that the "stick together" and have the same after-collision speed.

Set up your coordinate system so that particle A is moving in the positive direction along the x-axis (so its momentum is <2p, o>) and particle B is moving in the negative direction along the y-axis (so its momentum is <0, p>). Before the collision, their total momentum is <2p, p>. After the collision, the move off together: let's call their velocity vector <vx, vy>. Since their total mass is now 3m we have <2p, p>= <3mvx,3mvy> so we must have vx= 2p/(3m) and vy= p/(3m). v2 is now (3p/3m)2+ (p/3m)2= (10p2)/(9m2) and the kinetic energy is
(1/2)(3m)(10p2)/9m2)= (150/81)p2/m.

Of course, the total initial kinetic energy was still (9/4)p2/m so the energy lost is (9/4- 150/81)p2/m= ((720-600)/324)p2/m = (30/81)p2/m.

Does anyone else know. Thank you.

I just keep on ending up with the same answer.

~angel~ said:
I just keep on ending up with the same answer.

hmmm why did i get 3p^2/2m

That is what I got as the final KE for the second question, but subtracting it from the initial KE to get the KE lost gave me the same answer as the first queston.

Okay, we have established that the initial kinetic energy is

$K_1 = \frac{9p^2}{4m}.$

The initial x-component of momentum is $P_{1x} = 2p$ and the y-component is $P_{1y} = p$. After collision $P_{2x} = 3mv_x$ and $P_{2y} = 3mv_y$. Conservation of momentum states that

\begin{align*} \vec{P}_1 & = \vec{P}_2 \\ |\vec{P}_1| & = |\vec{P}_2| \\ 4p^2 + p^2 & = 9m^2v_x^2 + 9m^2v_y^2 \\ 5p^2 & = 9mv^2 \\ v ^2 = \frac{5p^2}{9m^2}. \end{align*}

Substituting into the kinetic energy equation you get

$K_2 = \frac{5p^2}{6m}$

and therefore

$K_\mathrm{lost} = \frac{17p^2}{12m}.$

## 1. What is a completely inelastic collision?

A completely inelastic collision is a type of collision in which two objects stick together after colliding and move as one mass. This means that the objects lose all of their kinetic energy and the collision is considered to be perfectly inelastic.

## 2. How is momentum conserved in a completely inelastic collision?

In a completely inelastic collision, momentum is conserved because the total momentum of the two objects before the collision is equal to the total momentum of the combined mass after the collision. This is due to the fact that the objects stick together and move with the same velocity after the collision.

## 3. What is the mathematical equation for a completely inelastic collision?

The mathematical equation for a completely inelastic collision is m1v1 + m2v2 = (m1 + m2)vf, where m represents mass, v represents velocity, and the subscripts 1 and 2 represent the two objects involved in the collision.

## 4. What are some real-life examples of completely inelastic collisions?

Some examples of completely inelastic collisions in everyday life include a car crashing into a wall, two clay balls colliding and sticking together, or a person catching a ball and bringing it to a stop. In all of these examples, the objects stick together and move as one mass after the collision.

## 5. How does the coefficient of restitution relate to completely inelastic collisions?

The coefficient of restitution is a measure of how much kinetic energy is conserved in a collision. In completely inelastic collisions, the coefficient of restitution is equal to 0 because all of the kinetic energy is lost. This means that the objects do not bounce off of each other and stick together instead.

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