Question

In: Physics

Consider the air tracks from lab: A car of mass m is set in motion with...

Consider the air tracks from lab: A car of mass m is set in motion with constant speed. As it moves down the track a mass of 2m is dropped onto the car without causing any damage. The car and additional load continue along the track. How does kinetic energy of the car + load compare to the original kinetic energy of the car?

1) The kinetic energy is 3 times the original kinetic energy

2) The kinetic energy is equal to the original kinetic energy

3) The kinetic energy is 1/2 times the original kinetic energy

4) The kinetic energy is 1/3 times the original kinetic energy

5) The kinetic energy is 2 times the original kinetic energy

Expert Solution

The momentum of the 2m mass is (2m ) u_o be fore collission with the car

here u_o is initial velocity of the 2m mass.

The momentum of the car before collission is 'mu' here u is the initial velocity of the car.

the total momentum before collission is (2m ) u_o + mu

After the collission 2m mass is along with the car without damage means it is 'inelastic collission'

Therefore the momentum of the system after collission is ( 2m + m ) V here 'V' is the final velocity of the system.

According to law of conservation of energy the total momentum before collission = the total momentum after collission

(2m ) u_o + mu = ( 2m + m ) V

2mu_o + mu = 3m V

2u_o +u = 3V since 'm ' get cancelled on both sides

V = ( 2u_o + u ) / 3 = ( 2u_o /3 ) + ( u / 3 )

since V is 1/3 of the initial velocity of car , u , the final kinetic energy of the system after collission is equal to the 1/3 of the original kinetic energy. [ Note : kinetic energy = ( 1/2 ) m v²

Hence the fourth option of the question is correct.

genius_generous answered 4 months ago

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