Consider the air tracks from lab: A car of mass m is set in motion with constant
ID: 1471165 • Letter: C
Question
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
Explanation / Answer
The momentum of the 2m mass is (2m ) uo be fore collission with the car
here uo 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 ) uo + 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 ) uo + mu = ( 2m + m ) V
2muo + mu = 3m V
2uo +u = 3V since 'm ' get cancelled on both sides
V = ( 2uo + u ) / 3 = ( 2uo /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.