How do I find the modulus of the average force a sphere gets from colliding with the ground?

Jun 2017
229
6
Lima, Peru
The problem is as follows:

A ball of $1\,kg$ in mass is thrown with a speed of $-10\vec{j}\,\frac{m}{s}$ from a height of $15\,m$ to a horizontal floor. Find the modulus of the average force in $N$ that the ball receives from the floor during the impact with the ground which lasts $0.1\,s$ and dissipates $150\,J$. (You may use the value of gravity $g=10\,\frac{m}{s^2}$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&290\,N\\
2.&300\,N\\
3.&310\,N\\
4.&320\,N\\
5.&330\,N\\
\end{array}$

This problem has left me go in circles as I don't know exactly how should I treat or use the information to obtain the average force?. I'm assuming that there is a conservation of momentum but as I mentioned I don't know what to do?. Can somebody help me here?.
 

skeeter

Math Team
Jul 2011
3,135
1,698
Texas
impulse equation ...

$F \Delta t = m\Delta v \implies F = \dfrac{m\Delta v}{\Delta t}$, where $F$ is the average force of impact changing momentum

you have mass and delta t, you need the change in velocity from impact with the ground

total initial mechanical energy = kinetic energy at impact

$mgh + \dfrac{1}{2}mv_0^2 = \dfrac{1}{2}mv_f^2 \implies v_f = -\sqrt{2gh + v_0^2} = -20 \text{ m/s}$

post impact, the velocity is positive and has 150J less energy ...

impact KE = $\dfrac{1}{2}m(-20)^2 = 200 \text{ J} \implies$ post impact KE = $50 \text{ J}$

bounce velocity $\dfrac{1}{2}mv_f^2 = 50 \implies v_f = 10 \text{ m/s}$

change in velocity from impact with the ground is $\Delta v = [10 - (-20)] \text{ m/s} = 30 \text{ m/s}$

$F = \dfrac{m\Delta v}{\Delta t} = 300 \text{ N}$
 
Jun 2017
229
6
Lima, Peru
impulse equation ...

$F \Delta t = m\Delta v \implies F = \dfrac{m\Delta v}{\Delta t}$, where $F$ is the average force of impact changing momentum

you have mass and delta t, you need the change in velocity from impact with the ground

total initial mechanical energy = kinetic energy at impact

$mgh + \dfrac{1}{2}mv_0^2 = \dfrac{1}{2}mv_f^2 \implies v_f = -\sqrt{2gh + v_0^2} = -20 \text{ m/s}$

post impact, the velocity is positive and has 150J less energy ...

impact KE = $\dfrac{1}{2}m(-20)^2 = 200 \text{ J} \implies$ post impact KE = $50 \text{ J}$

bounce velocity $\dfrac{1}{2}mv_f^2 = 50 \implies v_f = 10 \text{ m/s}$

change in velocity from impact with the ground is $\Delta v = [10 - (-20)] \text{ m/s} = 30 \text{ m/s}$

$F = \dfrac{m\Delta v}{\Delta t} = 300 \text{ N}$
All uses a pretty good logic but I'm wondering why the answers sheet states that the answer is $310\,N$?. Could it be that is there anything missing or overlooked during the analysis?. Can you check this please?. :)
 

skeeter

Math Team
Jul 2011
3,135
1,698
Texas
Maybe someone else can do the problem ...
 
Jun 2019
455
237
USA
*facepalm!*
The net force is equal to 300 N! But the earth is still imparting downward momentum on the ball at a rate of 10 N, so the floor must impart that much more upward momentum.
$F_{floor} = ma - F_g = \pm 300 ~N ~\vec{j} \pm 10 ~N ~\vec{j} = \pm 310 ~N ~\vec{j}$ (depending on whether $\vec{j}$ is up or down).

It took me about five minutes to realise that mistake.
Another feather in the cap of "Always, always, always draw your free-body diagram."