Question

One car has two and a half times the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9.0 m/s, they then have the same kinetic energy. What were the original speeds of the two cars

Answer 1

Answer:

$v_1 = 7.96 m/s$

$v_2 = 17.8 m/s$

Explanation:

Let the mass of the other car is "m" and its kinetic energy is

$K = \frac{1}{2}mv^2$

now the mass of the first car is two and half times and its kinetic energy is half that of other car

so we will have

$\frac{1}{2}(2.5m)v_1^2 = \frac{1}{2}(\frac{1}{2}mv^2)$

$2.5 v_1^2 = 0.5 v^2$

$v_1 = 0.447 v$

now speed of both cars is increased by value of 9 m/s

so now we will have same kinetic energy for both cars

$\frac{1}{2}(2.5 m)(0.447v + 9)^2 = \frac{1}{2}m(v + 9)^2$

$2.5(0.447 v + 9)^2 = (v + 9)^2$

$1.58(0.447v + 9) = v + 9$

$0.293v = 5.22$

$v = 17.8 m/s$

so speed of first car is

$v_1 = 0.447 v = 7.96 m/s$

$v_2 = 17.8 m/s$