One car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 7.0 m/s, then they have the same kinetic energy. What were the original speeds of the two cars?
Initial, K1 = 1/2 K2
Final, K1 = K2
m1 = 2m2
v1 = 1/2 v2
The Attempt at a Solution
Since K1 = K2 after the cars' velocities were increased by 7.0 m/s, I can write:
1/2 m1 (v1+7.0 m/s)^2 = 1/2 m2 (v2+7.0 m/s)^2
Since I know m1 = 2m2, I plug it in and I get
1/2 * 2m2 (v1+7.0m/s)^2 = 1/2 m2 (v2+7.0 m/s)^2 and then simplified:
m2 (v1 + 7.0 m/s)^2 = 1/2 m2 (v2 + 7.0 m/s)^2, and then simplified again:
2(v1 + 7.0 m/s)^2 = (v2 + 7.0 m/s)^2
I know that by the initial conditions given, 2v1 = v2 and I plug it in and get:
2(v1 + 7.0 m/s)^2 = (2v1 + 7.0 m/s)^2 and then I get rid of the squares and get:
Sqrt 2 (v1 + 7.0 m/s) = (2v1 + 7.0 m/s)
I start having trouble from here on. According to my teacher's notes, the step after the previous equation above given is v1 = 7.0 / sqrt 2, giving the answer of v1 = 4.9 seconds. Where did that come from?