Question

a ball of mass m = 1.60 kg is released from rest at a height h = 77.0 cm above a light vertical spring of force constant k as in figure a shown below. the ball strikes the top of the spring and compresses it a distance d = 8.00 cm as in figure b shown below. neglecting any energy losses during the collision, find the following. (a) find the speed of the ball just as it touches the spring. m/s (b) find the force constant of the spring. kn/m

Explanation:

Step1: Calculate speed when ball touches spring

Use free - fall kinematic equation $v^{2}=v_{0}^{2}+2gh$. Here $v_{0} = 0$ m/s, $g = 9.8$ m/s² and $h=77.0$ cm = 0.77 m.
$v=\sqrt{2gh}=\sqrt{2\times9.8\times0.77}$

Step2: Calculate force constant of spring

Use conservation of mechanical energy. Initial potential energy of ball $U_{i}=mg(h + d)$ is converted into elastic potential energy of spring $U_{s}=\frac{1}{2}kd^{2}$ at maximum compression. So $mg(h + d)=\frac{1}{2}kd^{2}$, and we can solve for $k=\frac{2mg(h + d)}{d^{2}}$. Given $m = 1.60$ kg, $h = 0.77$ m, $d=8.00$ cm = 0.08 m, $g = 9.8$ m/s².

Answer:

(a) $\sqrt{2\times9.8\times0.77}\approx3.88$ m/s
(b) $\frac{2\times1.6\times9.8\times(0.77 + 0.08)}{0.08^{2}}=\frac{2\times1.6\times9.8\times0.85}{0.0064}\approx4165$ N/m = 4.165 kN/m