The particles in the figure below undergo an elastic collision in one dimension.

Question

The particles in the figure below

undergo an elastic collision in one dimension. Their velocities before the collision are

Find the velocities of the two particles after the collision. (Indicate the direction with the sign of your answer.)

v1f = 1      m/s v2f = 2      m/s The particles in the figure below (m1 = 2.2 kg and m2 = 3.5 kg) undergo an elastic collision in one dimension. Their velocities before the collision are v1i = 12 m/s and v2i = ?7.5 m/s. Find the velocities of the two particles after the collision. (Indicate the direction with the sign of your answer.)

Explanation / Answer

Okay, so we know the following:

m1 = 2.2 kg

m2 = 3.5 kg

v1i = 12 m/s

v2i = -7.5 m/s

To find the final velocities of the two particles, we use the conservation of momentum equation:

(m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{1f}+m_{2}v_{2f})

Given the information about the two particles, we can't use the conservation of energy equation alone to find the final velocities. However, since the collision is elastic, kinetic energy is conserved so that we have the following relation:

(rac{1}{2}m_{1}v_{1i}^2+rac{1}{2}m_{2}v_{2i}^2=rac{1}{2}m_{1}v_{1f}^2+rac{1}{2}m_{2}v_{2f}^2)

This conservation of kinetic energy equation along with the conservation of momentum equation can be used to derive the following relation (I won't derive it here. It takes too much space):

(v_{1i}-v_{2i}=v_{2f}-v_{1f})

We can use this relation to eliminate either v1f or v2f in our conservation of momentum equation. I'll eliminate v2f by solving for it.

(v_{2f}=v_{1f}+v_{1i}-v_{2i})

Plug this into our momentum equation

(m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{1f}+m_{2}(v_{1f}+v_{1i}-v_{2i}))

Solve for v1f

(m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{1f}+m_{2}v_{1f}+m_{2}v_{1i}-m_{2}v_{2i})

(m_{1}v_{1i}-m_{2}v_{1i}+2m_{2}v_{2i}=(m_{1}+m_{2})v_{1f})

(v_{1f}=rac{(m_{1}-m_{2})v_{1i}+2m_{2}v_{2i}}{m_{1}+m_{2}})

This expression will give us the final velocity of the first particle. We can plug in the numbers

(v_{1f}=rac{(2.2 kg-3.5 kg)(12 m/s)+2(3.5 kg)(-7.5 m/s)}{2.2 kg+3.5 kg})

And the first particle's final velocity is:

(v_{1f}=-11.9 m/s)

Now, we can easily find the final velocity of the second particle by using that same relation I mentioned earlier:

(v_{2f}=v_{1f}+v_{1i}-v_{2i})

(v_{2f}=-11.9 m/s+12 m/s+7.5 m/s)

And the second particle's final velocity is:

(v_{2f}=7.6 m/s). The positive velocity indicates that it is moving to the right.

Whew! That was a long problem. But I hope this helps! :)

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