The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 6.0 m/s , they then have the same kinetic energy.

Question

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 6.0 m/sm/s , they then have the same kinetic energy.What were the original speeds of the two cars?

Express your answers using two significant figures separated by a comma.

Answer 1

Let m1 be the mass of the first car and m2 the mass of the second car; therefore m1 = 2 x m2.

The kinetic energy of the first car is 1/2 that of the second car or KE1 = 1/2 KE2.

KE1 = 1/2 x m1 x V1^2 which also equals 1/2 x 2 x m2 x V1^2.

KE2 = 1/2 x m2 x V2^2 where V1 and V2 are the initial velocities of the cars.

As KE1 = 1/2 KE2, we can write 1/2 x 2 x m2 x V1^2 = 1/2 x (1/2 x m2 x V2^2). After some simplification, this gives V1^2 = 1/4 x V2^2.

When both cars increase their speed by 2 m/s, the new velocities are v1new = V1 + 6 and V2new = V2 +6.

The new kinetic energies are equal; therefore, 1/2 x 2 x m2 x (V1 +6)^2 = 1/2 x m2 x (V2 + 6)^2.

Substitute for V1^2 in the previous equation, the value in terms of V2^2 given above, noting that V1 = √(1/4 x V2^2). Solve for V2. There is a lot of messy algebra that leads to V1 = 3√2, and V2 = 6√2.

Substitute V1 and V2 into the original KE equation (KE1 =1./2 x KE2).

KE1 = 1/2 x 2 m2 x (3√2)^2 = 18m2, KE2 = 1/2 x m2 x (6√2)^2 = 36m2; thus the initial velocities satisfy the initial kinetic energy relationship.