Answer to Question #266169 in Physics for Feyth

Question

a body of mass 0.1 kg drops from a height of 8m onto a hard floor bounces to a height of 2m. Calculate the momentum if the body is in contact with the floor for 0.1 seconds. What is the force exerted on the body (g=ms-²)

Answer 1

Let's first find the velocity of the body just before it hits a floor. Let's choose the downwards as positive direction. Then, we get:

$$v^2=v_0^2+2gh,$$ $$v=\sqrt{2gh}=\sqrt{2\times10\ \dfrac{m}{s^2}\times8\ m}=12.65\ \dfrac{m}{s}.$$

Then, we can find the initial velocity of the body when it bounces from the floor. Let's choose the upwards as the positive direction. Then, we get:

$$v^2=v_0^2+2gh,$$ $$0=v_0^2+2gh,$$ $$v_0=\sqrt{-2gh}=\sqrt{-2\times(-10\ \dfrac{m}{s^2})\times2\ m}=6.32\ \dfrac{m}{s}.$$

We can find the momentum as follows:

$$\Delta p=m(v_f-v_i),$$ $$\Delta p=0.1\ kg\times(6.32\ \dfrac{m}{s}-(-12.65\ \dfrac{m}{s}))=1.897\ \dfrac{kg\times m}{s}.$$

Finally, we can find the force exerted on the body:

$$F=\dfrac{\Delta p}{\Delta t}=\dfrac{1.897\ \dfrac{kg\times m}{s}}{0.1\ s}=18.97\ N.$$