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4-1.Newton's Laws of Motion
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A $40\ g$ ball dropped from a certain height bounces back from the horizontal ground without losing mechanical energy. If its speed is $10\ m/s$ just before making contact with the ground, and the average value of the force of the ground on the ball is equal to $16\ N$ while ball and wall are in contact, how long were they in contact .......... $s$
A$25$
B$50$
C$75$
D$100$
Solution
$\begin{gathered}
2 \times 40 \times {10^{ – 3}} \times 10 = 16\,t \hfill \\
50 \times {10^{ – 3}}\,s\, = \,t \hfill \\
\end{gathered} $
2 \times 40 \times {10^{ – 3}} \times 10 = 16\,t \hfill \\
50 \times {10^{ – 3}}\,s\, = \,t \hfill \\
\end{gathered} $
Standard 11
Physics
