Elastic Collision Between Different Masses


Consider two disks of equal size but different mass sliding on a smooth surface. Disk A has mass 250 grams and initial velocity of 2 m/s. Disk B has mass 175 grams and is initially at rest. The disks collide directly, with the velocity of Disk A along the line connecting their centers. The collision is totally elastic, with a restitution coefficient of 1.0. Determine the velocity of each disk immediately after the collision and show that the total kinetic energy is the same before and after the collision.


The coefficient of restitution, Cr, is the ratio of the velocity of separation to the velocity of approach in a collision. In the case of an elastic collision it is 1.0. For two moving objects, A and B, Cr = (vA-vB)/(vBi-vAi) where the i indicates initial conditions. Here is a good opportunity to get the signs messed up. The velocity of approach carries the opposite sign from the velocity of separation so the order of the terms in numerator and denominator must be reversed. In this case (vBi-vAi) = (vA-vB) since Cr=1.0. We also know that momentum is conserved so mA*vAi + mB*vBi = mA*vA + mB*vB. This gives us two equations in the two unknown post collision velocities, vA and vB.

Two solve this set of equations let's use the Cr relationship to get vA in terms of vB. We know that (0 - 2 m/s)=(vA-vB) or vA = vB-2. Plugging this into the momentum equation we get mA*vAi + mB*vBi = mA*(vB-2) + mB*vB = (mA + mB)*vB -2*mA. So vB = (mA*vAi+mB*vBi+2*mA)/(mA+mB). Putting in the known quantities we get vB = (.25*2+0+2*.25)/(.250+.175) = 1/.425 =
2.35294117647058823529411764705882 m/s.
Pardon the absurd number of decimal places but we are trying to make a point here.
Since vA=vB-2, vA=0.35294117647058823529411764705882/s

The kinetic energy of the system before the collision was all in disk A. KE=1/2*.25*22 Joules = 0.5 Joules. After the collision the kinetic energy is shared between the disks. The disk A kinetic energy, KEA is
1/2*.25*0.352941176470588235294117647058822 Joules =
0.0155709342560553633217993079584775 Joules.
KEB is 1/2*.175*2.352941176470588235294117647058822 Joules =
0.484429065743944636678200692041522 Joules.
The total KE after the collision is
0.499999999999999999999999999999522 Joules,
about as close to the original value as we might expect.

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