Question

Calculate expression for final velocity after elastic collison

Answer 1

When a collision between two objects is elastic, kinetic energy is conserved. In physics, the most basic way to look at elastic collisions is to examine how the collisions work along a straight line. If you run your bumper car into a friend’s bumper car along a straight line, you bounce off and kinetic energy is conserved. But the behavior of the cars depends on the mass of the objects involved in the elastic collision.

Here’s an example: You take your family to the Physics Amusement Park for a day of fun and calculation, and you decide to ride the bumper cars. You wave to your family as you speed your 300-kilogram car-and-driver up to 10.0 meters per second. Suddenly, Bonk! What happened? The 400-kilogram car-and-driver in front of you had come to a complete stop, and you rear-ended the car elastically; now you’re traveling backward and the other car is traveling forward. “Interesting,” you think. “I wonder if I can solve for the final velocities of both bumper cars.”

You know that the momentum was conserved, and you know that the car in front of you was stopped when you hit it, so if your car is Car 1 and the other is Car 2, you get the following:

m1vf1 + m2vf2 = m1vi1

However, this doesn’t tell you what vf1 and vf2are, because you have two unknowns and only one equation here. You can’t solve for vf1or vf2 exactly in this case, even if you know the masses and vi1. You need some other equations relating these quantities. How about using the conservation of kinetic energy? The collision was elastic, so kinetic energy was conserved. KE = (1/2)mv2, so here’s your equation for the two cars’ final and initial kinetic energies:

Now you have two equations and two unknowns, vf1 and vf2, which means you can solve for the unknowns in terms of the masses and vi1. You have to dig through a lot of algebra here because the second equation has many squared velocities, but when the dust settles, you get the following two equations:

Now you have vf1 and vf2 in terms of the masses and vi1. Plugging in the numbers gives you the two bumper cars’ final velocities. Here’s the velocity of your car:

And here’s the final velocity of the other guy:

The two speeds tell the whole story. You started off at 10.0 meters per second in a bumper car of 300 kilograms, and you hit a stationary bumper car of 400 kilograms in front of you. Assuming the collision took place directly and the second bumper car took off in the same direction you were going before the collision, you rebounded at –1.43 meters per second — backward, because this quantity is negative and the bumper car in front of you had more mass — and the bumper car in front of you took off at a speed of 8.57 meters per second.

Now you decide to go back and pick on some poor light cars in a monster bumper car. What happens if your bumper car (plus driver) has a mass of 400 kilograms and you rear-end a stationary 300-kilogram car? In this case, you use the equation for conservation of kinetic energy, the same formula you use in the previous example. Here’s what your final velocity comes out to:

The little car’s final velocity comes out to

In this case, you don’t bounce backward. The lighter, stationary car takes off after you hit it, but not all your forward momentum is transferred to the other car. Is momentum still conserved? Here are your formulas for the initial and final momentums:

pi = m1vi1

pf = m1vf1 + m2vf2

Putting in the numbers, here’s the initial momentum:

And here’s the final momentum:

The numbers match, so momentum is conserved in this collision, just as it is for your collision with a heavier car.

Answer 2

Answer:

If two particles are involved in an elastic collision, the velocity of the second particle after collision can be expressed as: v2f=2⋅m1(m2+m1)v1i+(m2−m1)(m2+m1)v2i v 2 f = 2 ⋅ m 1 ( m 2 + m 1 ) v 1 i + ( m 2 − m 1 ) ( m 2 + m 1 ) v 2 i