Question

A railroad car of mass m and speed v collides and sticks to an identical railroad car that is initially at rest. After the collision, the kinetic energy of the systemSelect one:A. is one fourth as much as before.B. is one third as much as before.C. is one quarter as much as before.D. is the same as before.E. is half as much as before.

Answer 1

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,

`mv=(m+m)v_0`

Where v₀ is the velocity of the two cars that are stuck together after the collision.

On simplifying,

`\begin{gathered} mv=2mv_0 \\ \Rightarrow v_0=(v)/(2) \end{gathered}`

The kinetic energy before the collision is,

`K_1=(1)/(2)mv^2`

The kinetic energy before the collision is,

`\begin{gathered} K_2=(1)/(2)mv^2_0 \\ =(1)/(2)m((v)/(2))^2 \\ =(1)/(2)(mv^2)/(4) \\ =(K_1)/(4) \end{gathered}`

Thus the kinetic energy of the system after the collision is one-fourth as much as before.

Threfore, the correct answer is option A.