Question

10 kg ball rolls off a 5 m high cliff traveling at 2 m/s. if it lands on a spring with a spring constant of 10,000 N/m, how far can it compress the spring? (assume the spring is pointed in the direction of the ball)

Answer 1

Approximately
`0.3\; {\rm m}`. (Assuming that
`g = 9.81\; {\rm N\cdot kg^(-1)}`.)

Step-by-step explanation:

In this question, the ball initially possesses gravitational potential energy
`\rm GPE}` and kinetic energy
`{\rm KE}`. That energy was converted into the elastic potential energy
`{\rm EPE}` of the compressed spring.

Let
`m` denote the mass of the ball. When the height of the ball changes by
`\Delta h`, the change in the
`{\rm GPE}` of the ball would be
`{\rm GPE} = m\, g\, \Delta h`.

Let
`v` denote the initial speed of the ball. The initial kinetic energy of the ball would be
`{\rm KE} = (1/2)\, m\, v^(2)`.

Assume that the height of the cliff far exceeds the height of the spring. Thus, the change in the height of the ball would be approximately the same as the height of the cliff:
`\Delta h \approx 5\; {\rm m}`. The
`{\rm GPE}` of the ball would be:

`\begin{aligned}{\rm GPE} &= m\, g\, \Delta h \\ &\approx (10)\, (9.81)\, (5)\; {\rm J} \\ &= 490.5\; {\rm J} \end{aligned}`.

With a speed of
`v = 2\; {\rm m\cdot s^(-1)}`, the initial
`{\rm KE}` of the ball would be:

`\begin{aligned}{\rm KE} &= (1)/(2)\, m\, v^(2) \\ &= (1)/(2)\, (10)\, (2)^(2)\; {\rm J} \\ &= 20\; {\rm J}\end{aligned}`.

Let
`k` denote the spring constant of the spring. With a displacement of
`x`, the
`{\rm EPE}` in the spring would be
`{\rm EPE} = (1/2)\, k\, x^(2)`.

All that
`{\rm GPE} + {\rm KE} \approx (490.5 + 20)\; {\rm J} = 510.5\; {\rm J}` of energy would have been converted into the
`{\rm EPE}` of the spring.

It is given that
`k = 10000\; {\rm N\cdot m^(-1)}`. In other words, when
`x` is in meters:

`(1/2)\, (10000)\, x^(2) = {\rm EPE} \approx 510.5`.

Solve for the displacement of the spring,
`x`:

`\begin{aligned} x &\approx \sqrt{(510.5)/((1/2)\, (10000))} \approx 0.3\; {\rm m}\end{aligned}`.