What were the original speeds of the cars?

  • Thread starter Libohove90
  • Start date
1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Initial, K1 = 1/2 K2

Final, K1 = K2

m1 = 2m2

K=1/2 mv^2

v1 = 1/2 v2

3. 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?


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Distribute the sqrt(2) in the left hand side over both terms. Then collect like terms on each side of the equation (so that everything that multiplies v1 is on one side, and everything that multiplies 7 m/s is on the other side). Then, solve for v1. On the other side of the equation, you'll have a fractional expression multiplying 7 m/s. Rationalize the denominator of this expression and simplify.

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