Question

A ball of mass $1kg$ dropped from $9.8m$ height strikes the ground and rebounds to a height of $4.9m$. If the time of contact between ball and ground is $0.1s$, then find impulse and average force acting on the ball.

Answer 1

Step 1: Given Data

Mass of the ball $m=1kg$

The initial velocity of the ball $u=0$

Let the final velocity when it just strikes the ground be $v_1$.

Let the initial velocity for rebound be $v_2$.

Let the first height be $h_1=9.8m$

Let the second height after collision be $h_2=4.9m$

Let the acceleration due to gravity be $g=9.8m/s^2$

Change in time $∆t=0.1s$

Step 2: Calculate final velocity when it just strikes the ground

For the first case, according to Newton's third law of motion

$v^2=u^2+2aS$

$\Rightarrow {v_1}^2=0+2g{h}_1$

$\Rightarrow {v_1}^2=2\times 9.8\times 9.8$

$\Rightarrow v_1=9.8\sqrt{2}m/s$

Step 3: Calculate initial velocity for rebound

For the rebound case, the final velocity will get zero, and acceleration due to gravity will be negative for upward motion.

According to Newton's third law of motion

$v^2=u^2+2aS$

$\Rightarrow 0=v_2-2g{h}_2$

$\Rightarrow {v_2}^2=2\times 9.8\times 4.9$

$\Rightarrow v_2=\sqrt{2\times 2\times 4.9\times 4.9}$

$\Rightarrow v_2=2\times 4.9=9.8m/s$

Step 4: Calculate the Impulse

Therefore the initial moment is

$\overrightarrow{p_i}=-m{v}_1\hat{j}$

Therefore the final moment is

$\overrightarrow{p_2}=m{v}_2\hat{j}$

We know that impulse is equal to a change in momentum.

$\therefore ∆\overrightarrow{p}=\overrightarrow{p_f}-\overrightarrow{p_i}$

$=m{v}_2\hat{j}-\left(-m{v}_1\hat{j}\right)$

$=m\left(v_2+v_1\right)\hat{j}$

$=1\left(9.8\sqrt{2}+9.8\right)\hat{j}$

$$\begin{gathered} =9.8\left(\sqrt{2}+1\right)\hat{j} \\ =9.8\times 2.4\hat{j} \\ =23.52\hat{j}Ns \end{gathered}$$

Step 5: Calculate the average force

We know that impulse can be given as,

$I=\int Fdt=∆p$

For average force, we can write

$F_{avg}=\frac{∆p}{∆t}$

$=\frac{23.52}{0.1}=235.2N$

Hence, the impulse is $23.52Ns$ in the positive $y$ direction, and the average force acting on the ball is $235.2N$.