Answer to Question #156592 in Physics for KIm

Question

Aubrey drops a ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.960 m. What impulse was given to the ball by the floor.

Answer 1

By definition, the impulse given to the ball is equal to the change of its momentum:

$$J = \Delta p = m(v_2 - v_1)$$

where $m = 0.15kg$ is the mass of the ball, $v_1, v_2$ are the projections of the ball's velocity on the vertical axis before and after the collision respectively.

Let's direct the vertical axis upwnward.

According to the mechanical energy conservation, the kinetic energy of the ball right before the collision is equal to its potential energy at the begining of the motion. Writting down the corresponding formulas for these energies, obtian:

$$\dfrac{mv_1^2}{2} = mgh_1$$

where $g = 9.81m/s^2$ is the gravitational acceleration, and $h_1 =1.25m$ is the initial height. Thus, obtain:

$$v_1 = -\sqrt{2gh_1}$$

(negative, since before the collision the velocity is directed downward).

After the collision, the kinetic energy of the ball is equal to its potential energy at the maximum height $h_2 = 0.96m$. Obtain:

$$\dfrac{mv_2^2}{2} = mgh_2$$ $$v_2 = \sqrt{2gh_2}$$

(positive, since after the collision the velocity is directed upward).

Finally, obtain the expression for the impulse:

$$J = m(v_2 - v_1) = m\sqrt{2g}(\sqrt{h_1} + \sqrt{h_1})\\ J = 0.15\cdot\sqrt{2\cdot 9.81}\cdot (\sqrt{1.25} + \sqrt{0.96}) \approx 1.39 N\cdot s$$

Answer. 1.39 N*s.