Question

Q2. One car has twice the mass of a second car, but only 1/3 as much kinetic energy. When both cars increase their speed by 6 m/s they then have the same kinetic energy. What were the original speeds of the two cars?

Answer 1

Suppose mass of car 1 is m1, and mass of car 2 is m2, then

m1 = 2*m2

kinetic energy of m1 is K1, and kinetic energy of m2 is K2, then

K1 = (1/3)*K2

speed of car 1 is V1, and speed of car 2 is V2, then

Since Kinetic energy, K = (1/2)*m*V^2, So

(1/2)*m1*V1^2 = (1/3)*(1/2)*m2*V2^2

Since m1 = 2*m2, So

(1/2)*2*m2*V1^2 = (1/3)*(1/2)*m2*V2^2

2*V1^2 = (1/3)*V2^2

V2^2 = 6*V1^2

V2 = V1*sqrt 6

Now given that speed of both cars is increased by 6 m/s, After that

K1' = K2'

(1/2)*m1*(V1 + 6)^2 = (1/2)*m2*(V2 + 6)^2

2*m2*(V1 + 6)^2 = m2*(V2 + 6)^2

2*(V1 + 6)^2 = (V2 + 6)^2

Since, V2 = V1*sqrt 6, So

2*(V1 + 6)^2 = (V1*(sqrt 6) + 6)^2

2*V1^2 + 72 + 24*V1 = 6*V1^2 + 36 + V1*12*sqrt 6

4*V1^2 + V1*(12*sqrt 6 - 24) - 36 = 0

V1^2 + V1*(3*sqrt 6 - 6) - 9 = 0

V1^2 + 1.3485*V1 - 9 = 0

Now solving above quadratic equation:

V1 = [-1.3485 +/- sqrt (1.3485^2 - 4*1*(-9))]/2

taking +ve root only, since speed is positive

V1 = [-1.3485 + sqrt (1.3485^2 + 36)]/2

V1 = 2.40 m/s = original speed of car 1

V2 = V1*sqrt 6 = 2.40*sqrt 6

V2 = 5.88 m/s = original speed of car 2

Let me know if you've any query.