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?.

Supposedly the answer is $310\,N$. But the only thing that comes to my mind to find the average force is:

$\overline{\vec{F}}=m\frac{\Delta v}{\Delta t}$

Since there will be a conservation of mechanical energy at the impact it can be found using this analysis. But again I'm not sure how to get to the answer of $310\,N$. Can somebody help me here?.

  • $\begingroup$ So all you need is calculate $\Delta v=v_1-v_0$ where $\frac{mv_0^2}{2}=\frac{m(-10j\frac{m}{s})^2}{2}+mgh$ and $\frac{mv_0^2}{2}-150J=\frac{mv_1^2}{2}$ and $v_1,v_0$ have different signs. $|v_1|=10\frac{m}{s}$, ... $\endgroup$ – Alexey Burdin Dec 2 '19 at 11:24

Velocity before impact = $-20j$ m/s

Energy before impact = $0.5*(1)(20)^2 = 200$ J

Energy after impact = $50$ J

Velocity after impact = $10j $ m/s

Average Net Force = $\frac{10(1)- (-20)(1) }{0.1} = 300j $ N

Average Force of floor = Average Net force - Average Body Force

Hence average force of floor = $300 - (-10) = 310 N$


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