ap physics 1 momentum

Question

She makes it out the mall into the parking structure and sees the security’s golf cart. After realizing her stroke of luck, she takes off away from the hustle and bustle of the holiday shopping. She coasts from rest down the 30.0 m tall ramp to exit the parking structure in the 350. kg cart (including clerk). She hits a 200. kg concrete trashcan that was already knocked over, at the bottom, rebounds off it at 10.0 m/s and starts the can rolling. If the trash can then rolls down another 30.0 meter tall incline, what will be the speed of the can at the bottom?

mass of "she" /clerk = 55kg

Answer 1

This problem involves concepts of potential and kinetic energy in addition to conservation of momentum. Let's follow the events here chronologically from beginning to end.

First, the clerk and cart, with combined mass of 350 kg, roll down a ramp with 30 meter elevation. We can find how much gravitational potential energy was released here with the expression m*g*h, where m is the mass of the object, g is the acceleration due to gravity, and h is the elevation change. This is equal to 350 kg * 9.81 m/s² * 30 m = 103,005 joules.

Now, what actually happened to all that potential energy? It became kinetic energy, of course. We can thus equate this quantity of 103,005 J to the expression for kinetic energy of a moving object, 1/2 * m * v², where m is still the mass and v is the velocity of the object. Solving for v, we find that the velocity of the cart (and the clerk inside) at the bottom of the ramp is approximately 24.261 m/s. (I hope she's wearing a helmet!)

Now we're ready to consider the collision with the concrete trash can. This is where momentum gets involved. The change in momentum (a.k.a. impulse) that the cart experiences during the collision is equal to m*Δv, the mass times the change in velocity. Since it comes into the collision moving forward at 24.261 m/s and exits the collision at 10.0 m/s, the impulse is 350 kg * 14.261 m/s = 4,991.38 kg*m/s. (We know that the cart's velocity is still directed forward because its mass is greater than that of the trash can.)

Due to conservation of momentum, we know that the same quantity of impulse must be imparted to the trash can. Dividing by its mass of 200 kg, we can calculate that its velocity after the collision must be 24.96 m/s. Once again using the formula 1/2 * m * v², we find that its kinetic energy is 62,284.64 J.

Now the trash can rolls down its own ramp, and once again we're adding kinetic energy equal to the released gravitational potential energy m*g*h. 200 kg * 9.81 m/s² * 30 m = 58,860 J, so adding this on to our previous kinetic energy, we have a final value of 121,144.64 J of kinetic energy for the trash can at the bottom of the ramp. Equating to the kinetic energy formula 1/2 * m * v² one final time to solve for velocity, we find that the velocity of the can is thus 34.8 m/s. Yikes! Let's hope it didn't flatten any pedestrians.