Question

The weight of the steel ball is 4.1 g. It was dropped from a height of 2.0 m into the sand. The ball sank into the sand so that the depth of the resulting pit was 27 mm. How much average force did the sand apply to the ball?

Answer 1

Let's first find the velocity of the ball near the ground from the law of conservation of energy:

$$PE=KE,$$ $$mgh=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{2gh}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2} \cdot2\ m}=6.3\ \dfrac{m}{s}.$$

Let's find the time that the ball takes to fully stops when it sank into the sand:

$$t=\dfrac{s}{v}=\dfrac{27\cdot10^{-3}\ m}{6.3\ \dfrac{m}{s}}=4.3\cdot10^{-3}\ s.$$

Finally, we can find the average force that the the sand apply to the ball from impulse-momentum change equation:

$$F_{avg}\Delta t=m\Delta v,$$ $$F_{avg}\Delta t=m(v_f-v_i),$$ $$F_{avg}=\dfrac{m(v_f-v_i)}{\Delta t},$$ $$F_{avg}=\dfrac{4.1\cdot10^{-3}\ kg\cdot(0-6.3\ \dfrac{m}{s})}{4.3\cdot10^{-3}}=-6.0\ N.$$

The magnitude of the average force that the sand apply to the ball equals 6.0 N. The sign minus means that the average force directed upward in the opposite direction to the motion of the ball.

Answer:

$F_{avg}=6.0\ N,$ upward.