Question

A runaway railroad car, with mass 30x10^4 kg, coasts across a level track at 2.0 m/s when it collides with a spring loaded bumper at the end of the track. If the spring constant of the bumper is 2x10^6 N/m, what is the maximum compression of the spring during the collision? (Assume collision is elastic)

Answer 1

0.6 m

Step-by-step explanation:

The law of conservation of energy states that:

`\Delta E_m=0`

The mechanical energy (
`E_m`) is the sum of the kinetic energy and the potential energy:

`\Delta K+\Delta U=0\\K_f-K_i+U_f-U_i=0\\(mv_f^2)/(2)-(mv_i^2)/(2)+(kx_f^2)/(2)-(kx_i^2)/(2)=0`

`U_i` is zero since the spring is not initially compressed and
`K_f` is zero since all kinetic energy becomes potentital energy:

`(kx_f^2)/(2)=(mv_i^2)/(2)`

Finally, we solve for x and replace the given values:

`x_f^2=(mv_i^2)/(k)\\x_f=\sqrt{(mv_i^2)/(k)}\\x_f=\sqrt{((30*10^4kg)(2(m)/(s))^2)/(2*10^6(N)/(m))}\\x_f=0.6 m`

Answer 2

0.775 m

Step-by-step explanation:

As the car collides with the bumper, all the kinetic energy of the car (K) is converted into elastic potential energy of the bumper (U):

`U=K\\frac{1}{2}kx^2 = (1)/(2)mv^2`

where we have

`k=2\cdot 10^6 N/m` is the spring constant of the bumper

x is the maximum compression of the bumper

`m=30\cdot 10^4 kg` is the mass of the car

`v=2.0 m/s` is the speed of the car

Solving for x, we find the maximum compression of the spring:

`x=\sqrt{(mv^2)/(k)}=\sqrt{((30\cdot 10^4 kg)(2.0 m/s)^2)/(2\cdot 10^6 N/m)}=0.775 m`