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

First car will use subscript 1 and second car will use subscript 2

m₁=2*m₂

E₁=0.5*E₂

E=0.5*m*v²

v and E will have a subscript of O for Original Speed and F for Final Speed

v_1F=v_1O+6.0m/s

v_2F=v_2O+6.0m/s

E_1F=E_2F

Find v_1O&v_2O

0.5*m₁*v_1F²=0.5m₂*v_2F²

0.5*m₁*v_1F²=0.5m₂*v_2F²

m₁*v_1F²=m₂*v_2F²

Substitute 2m₂ for m₁

2*m₂*v_1F²=m₂*v_2F²

2*m₂*v_1F²=m₂*v_2F²

2*v_1F²=v_2F²

Take Square Root of both sides abbreviated here as sqrt

sqrt(2)*v_1F=v_2F

Substitute the following equation in for v_1F=v_1O+6.0m/s

sqrt(2)*(v_1O+6.0m/s)=v_2F

Substitute the following equation in for v_2F=v_2O+6.0m/s

sqrt(2)*(v_1O+6.0m/s)=v_2O+6.0m/s

Isolate v_2O

sqrt(2)*(v_1O+6.0m/s)-6.0m/s=v_2O

Use E_1O=0.5E_2O to find original speed relationship

0.5*m₁*v_1O²=0.5*0.5*m₂*v_2O²

Remove the common 0.5 term

0.5*m₁*v_1O²=0.5*0.5*m₂*v_2O²

m₁*v_1O²=0.5*m₂*v_2O²

Use m₁=2*m₂ to remove mass relationship

2*m₂*v_1O²=0.5*m₂*v_2O²

2*m₂*v_1O²=0.5*m₂*v_2O²

2*v_1O²=0.5*v_2O²

Isolate v_2O

4*v_1O²=v_2O²

Take sqrt of both sides of equation

2*v_1O=v_2O

Use previous equation sqrt(2)*(v_1O+6.0m/s)-6.0m/s=v_2O to substitute in for v_2O

sqrt(2)*(v_1O+6.0m/s)-6.0m/s=2*v_1O

1.4v_1O+8.5m/s-6.0m/s=2*v_1O

Combine constants and move v_1O to right side of equation

2.5m/s=0.6v_1O

Solve for v_1O

v_1O=2.5m/s*0.6=1.5m/s

Solve for v_2O

v_2O=2*v_1O

v_2O=2*1.5m/s=2.9m/s

Note:Rounding occurs after math is performed v_1O is actually 1.45m/s before rounding.

Answer:

v_1O=1.5m/s

v_2O=2.9m/s