Given,
The mass of the car is m
The intial velocity of the car is v
From the law of conservation of momentum, the momentum of the system remains constant.
Thus,
[tex]mv=(m+m)v_0[/tex]Where v₀ is the velocity of the two cars that are stuck together after the collision.
On simplifying,
[tex]\begin{gathered} mv=2mv_0 \\ \Rightarrow v_0=\frac{v}{2} \end{gathered}[/tex]The kinetic energy before the collision is,
[tex]K_1=\frac{1}{2}mv^2[/tex]The kinetic energy before the collision is,
[tex]\begin{gathered} K_2=\frac{1}{2}mv^2_0 \\ =\frac{1}{2}m(\frac{v}{2})^2 \\ =\frac{1}{2}\frac{mv^2}{4} \\ =\frac{K_1}{4} \end{gathered}[/tex]Thus the kinetic energy of the system after the collision is one-fourth as much as before.
Threfore, the correct answer is option A.
