Question

a) Show that three concurrent forces F1, F2 and F3 will be in equilibrium when the resultant of F1 and F2 is equal and opposite to third force F3.
b) A ball of mass $m$, falls from a height of $H$, and rebounds back to a height $h$. Find the impulse, and the average force, between the ball and the ground, if the time of contact is $\Delta t$.

Answer 1

In the first part, we shall analyze the properties of forces acting on a body and then look forward to the situation when a body is in equilibrium even when multiple forces are acting on it. In the second part, we shall find the impulse of a ball falling and colliding with the ground and as a result of which the velocity of the ball changes.

Complete step-by-step solution:
(a)
If a body is in equilibrium even if multiple forces are acting on it, then the vector sum (as force is a vector quantity) of all those forces must be zero and the forces must have balanced each other.
$\begin{aligned} & \vec{F}1+\vec{F}2+\vec{F}3=0 \\\ & \Rightarrow \vec{F}1+\vec{F}2=-\vec{F}3 \\\ \end{aligned}$

Therefore, the resultant of F1 and F2 is equal and opposite to the third force F3.

(b)
When the ball hits the ground, velocity, $v=\sqrt{2gH}$
Where,
$g=$acceleration due to gravity
$H=$initial height of ball
After falling onto the ground, it rebounds to a height of $h,$
Thus, new velocity of ball, $v'=\sqrt{2gh}$
Impulse is defined as the change in momentum per unit time.

Here, the change in momentum, $\Delta p$ = final momentum – initial momentum
$\Delta p={{m}_{2}}{{v}_{2}}-{{m}_{1}}{{v}_{1}}$
Where,
${{m}_{1}},{{v}_{1}}=$ initial mass and velocity of ball
${{m}_{2}},{{v}_{2}}=$final mass and velocity of ball
$\begin{aligned} & \Rightarrow \Delta p=mv'-mv \\\ & \Rightarrow \Delta p=m\left( \sqrt{2gh} \right)-m\left( \sqrt{2gH} \right) \\\ \end{aligned}$
Hence, impulse, $J=\dfrac{\Delta p}{\Delta t}$
$\Rightarrow J=\dfrac{m\left( \sqrt{2gh} \right)-m\left( \sqrt{2gH} \right)}{\Delta t}$
$\Rightarrow J=\dfrac{m\left( \sqrt{2gh}-\sqrt{2gH} \right)}{\Delta t}$ ………………………. Equation (1)

Average force is given by multiplying mass with average acceleration.
$\Rightarrow {{F}_{average}}=m{{a}_{average}}$

Now, average acceleration is total velocity divided by total time taken.
$\Rightarrow {{a}_{average}}=\dfrac{\Delta v}{\Delta t}$
Here,
$\begin{aligned} & \Delta v=v'-v \\\ & \Rightarrow \Delta v=\sqrt{2gh}-\sqrt{2gH} \\\ \end{aligned}$
$\Rightarrow {{a}_{average}}=\dfrac{\sqrt{2gh}-\sqrt{2gH}}{\Delta t}$
$\Rightarrow {{F}_{average}}=m\left( \dfrac{\sqrt{2gh}-\sqrt{2gH}}{\Delta t} \right)$
$\Rightarrow {{F}_{average}}=\dfrac{m\left( \sqrt{2gh}-\sqrt{2gH} \right)}{\Delta t}$ …………………….. equation (2)

Moreover, from equations (1) and (2), we see that impulse is also equal to the average force between two bodies, (here they are the ball and the ground)

Note:
When the ball is at a height, $H$, it has only potential energy stored in it due to its position and has zero kinetic energy. Therefore, the total energy in the ball is the gravitational potential energy $=mgH$. When the ball falls towards the ground, the potential energy gradually changes to kinetic energy due to motion of the ball as the ball acquires some velocity. Thus, $\dfrac{1}{2}m{{v}^{2}}=mgH$, or $v=\sqrt{2gH}$.