Answer
The percentage of the original kinetic energy that was lost in the collision is $~~33.3\%$
Work Step by Step
Let $v_0$ be the speed of the car before the collision.
We can find an expression for the initial kinetic energy:
$K_i = \frac{1}{2}mv_0^2$
$K_i = \frac{1}{2}(1000)v_0^2$
$K_i = 500~v_0^2$
We can use conservation of momentum to find an expression for the speed after the collision:
$p_f = p_i$
$(1500~kg)~v_f = (1000~kg)~v_0$
$v_f = \frac{(1000~kg)~v_0}{1500~kg}$
$v_f = \frac{2}{3}~v_0$
We can find an expression for the final kinetic energy:
$K_f = \frac{1}{2}mv_f^2$
$K_f = \frac{1}{2}(1500)(\frac{2}{3}~v_0)^2$
$K_f = \frac{1000}{3}~v_0^2$
We can find the percentage of the original kinetic energy that was lost:
$\frac{K_i-K_f}{K_i}\times 100\% =\frac{500~v_0^2-\frac{1000}{3}~v_0^2}{500~v_0^2}\times 100\% = 33.3\%$
The percentage of the original kinetic energy that was lost in the collision is $~~33.3\%$
