Question

a ball of mass 2.0 kg is released from rest at a height 60 cm above a light vertical spring of force constant k. the ball strikes the top of the spring and it compresses it a distance of 10 cm. neglecting any energy loss during the collision, find the spped of the ball just as it touches the spring? the force constant k of the spring?

Answer 1

A ball of mass $2.0\,\mathrm{kg}$ is released from rest at a height $60\,\mathrm{cm}$ above a light vertical spring of force constant $k$. The ball strikes the top of the spring and it compresses it a distance of $10\,\mathrm{cm}$. Neglecting any energy loss during the collision, find the speed of the ball just as it touches the spring? The force constant $k$ of the spring?

Solution:

Let:

$$m = 2.0\,\mathrm{kg}$$ $$h = 60\,\mathrm{cm} = 0.6\,\mathrm{m}$$ $$x = 10\,\mathrm{cm} = 0.1\,\mathrm{m}$$

$v - ?, k - ?$

According to the law of conservation energy the changing of potential energy of the ball is equal to changing of potential energy of the string:

$$\Delta E = \Delta W =>$$

$mg\Delta H = \frac{1}{2} kx^2$, where $g$ is the acceleration due to gravity.

$$mg(h + x) = \frac{1}{2} kx^2$$ $$k = \frac{2mg(h + x)}{x^2}$$ $$k = \frac{2*9.8*(0.6 + 0.1)}{0.1^2} = 1372\,\mathrm{N/m}$$

From: $h = \frac{1}{2} gt^2$, $v = gt$

$$h = \frac{v^2}{2g} => v = \sqrt{2gh}$$ $$v = \sqrt{2*9.8*0.6} = 3.43\,\mathrm{m/s}$$

Answer: $v = 3.43\,\mathrm{m/s}$, $k = 1372\,\mathrm{N/m}$.