A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a
level (frictionless) surface and collides with a light spring-loaded guardrail with a spring constant
k of 1.0 x 106 N/m. Find the maximum distance the spring is compressed.

0.51 m

Explanation:

Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.

Kinetic energy, KE=½kx²

Where k is spring constant and x is the compression of spring

Potential energy, PE=mgh

Where g is acceleration due to gravity, h is height and m is mass

Equating KE=PE

mgh=½kx²

Making x the subject of formula

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s