Question :

A 2.0 kg mud ball drops from rest at a height of 15 m. If the impact between the ball and the ground lasts 0.50 s, what is the average force exerted by the ball on the ground


I have this setup :

enter image description here

Taking gravity $ g = 10 $.

At $A$ potential energy $ = mgh = 15(10)(2) = 300$.

Using Conservation of energy as:

$$ W + PE_0 + KE_0 = PE_f + KE_f + \text{Energy(Lost)} $$

Where all energies other than $PE$ and $KE$ are zero gives

$$ PE_A = KE_B $$

So that the kinetic energy when impact starts is equal to the initial potential energy, which is $300$.

From this we can find the velocity as

$$ KE = \frac{1}{2}mv^2 $$


$$ v = \sqrt{300} \approx 17.32 $$

Using Impulse momentum theorem we have

$$ I = F \Delta t = \Delta p = m (v_1 - v_0) $$

Here $v_0 = 0$ and $v_1$ has been found as $\sqrt{300}$.

From this we have

$$ F \Delta t = m(\sqrt{300}) $$


$$ F = \frac{m\sqrt{300}}{\Delta t} $$

Here $\Delta t = 0.5$ then the average force exerted is $F = 4\sqrt{300} = 40 \sqrt{3} \approx 69.28$

Therefore the average force exerted is $69.28$ to 2 decimal places


closed as off-topic by ACuriousMind Apr 29 '17 at 18:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ So what is your question? $\endgroup$ – Farcher Apr 29 '17 at 18:13
  • $\begingroup$ @Farcher the question is stated at the top. My working is what I'm not confident with and I don't have any solutions. $\endgroup$ – baxx Apr 29 '17 at 18:15

you can find the velocity at the bottom of the height. That comes out to be √(300) .also note that this would also be the velocity with which the ball bounces back up . so if you want to find the force exerted by the ball on the ground. Find the change in velocity and apply F =m (v1-v2)/t The answer would be double of what you got or 138.56

  • $\begingroup$ I used that formula though? I thought so at least. At which part are you referring to find the change in velocity at? $\endgroup$ – baxx Apr 29 '17 at 19:05
  • 1
    $\begingroup$ When it is just about to hit the ground its velocity is √(300) downward and later its √(300) upward. Thus change in velocity would be 2*√(300) $\endgroup$ – Lakshya Gupta Apr 29 '17 at 19:10
  • $\begingroup$ Thanks - I've considered the initial velocity as zero , but you're right, I should have considered the velocity change from before to after the impact. I'll try and write it up nicely. $\endgroup$ – baxx Apr 29 '17 at 19:15

Not the answer you're looking for? Browse other questions tagged or ask your own question.