# Homework Help: Symbolic Kenetic Energy

1. Feb 6, 2010

### starwars89625

1. The problem statement, all variables and given/known data
One car has one and a half times the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 8.0 m/s, they then have the same kinetic energy. What were the original speeds of the two cars?

2. Relevant equations
ke = (1/2)mv^2

3. The attempt at a solution
m1 = 1.5m2
k1 = k2/2
(1/2)(1.5m)(v1+8)^2 = (1/2)(m)(v2+9)^2

2. Feb 6, 2010

### OmCheeto

This question makes my brain hurt.

ke1 = 1/2*m1*v1^2 = 1/4*m2*v2^2 = k2/2
getting rid of the fractions yields:
2m1*v1^2 = m2*v2^2
3m2*v1^2 = m2*v2^2
the m2's cancel leaving
3v1^2 = v2^2
so
v2 = ?

replacing v2 in your 3rd equation should yield one easily solved quadratic equation.
(1/2)(1.5m)(v1+8)^2 = (1/2)(m)(v2+8)^2

3. Feb 6, 2010

### Spinnor

I think that is it, except (v2+9) should be (v2+8). Multiply out all the terms and solve via the quadratic equation, see:

4. Feb 6, 2010

### downwithsocks

Before their speeds increase:
KE of car 1 = 1/2(Ke of car 2)
.5*1.5m*v1^2 = .5(.5*m*v2^2)
.75mv1^2 = .25mv2^2
v2^2 = 3v1^2
v2 = 1.73v1

After the increase:
KE1 = KE2
.5*1.5m*(v1 + 8)^2 = .5*m*(v2 + 8)^2
.75m(v1 + 8)^2 = .5m(v2 + 8)^2
.75m(v1 + 8)^2 = .5m(1.73v1 + 8)^2

Hope I did that right, I'm kind of tired. There might be an easier way too, but you should be able to go from there.