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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


working

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

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

Meaning

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

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

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

And

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

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

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

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  • 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
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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

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  • $\begingroup$ 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
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    $\begingroup$ 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

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