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4-1.Newton's Laws of Motion
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A $150\, g$ tennis ball coming at a speed of $40\, m/s$ is hit straight back by a bat to a speed of $60\, m/s$. The magnitude of the average force $F$ on the ball, when it is in contact for $5\, ms$, is ........... $N$
A$2500$
B$3000$
C$3500$
D$4000$
(AIIMS-2011)
Solution
$\begin{array}{l}
The\,change\,in\,momentum\,\\
\Delta p = m\left( {{v_f} – {v_i}} \right)\\
\,\,\,\,\,\,\, = 0.150(60 – ( – 40)]\\
\,\,\,\,\,\,\, = 0.150 \times 100 = 15Ns\\
Thus,\,F = \frac{{\Delta p}}{{\Delta t}} = \frac{{15}}{{5 \times {{10}^{ – 3}}}} = 3 \times {10^3}N
\end{array}$
The\,change\,in\,momentum\,\\
\Delta p = m\left( {{v_f} – {v_i}} \right)\\
\,\,\,\,\,\,\, = 0.150(60 – ( – 40)]\\
\,\,\,\,\,\,\, = 0.150 \times 100 = 15Ns\\
Thus,\,F = \frac{{\Delta p}}{{\Delta t}} = \frac{{15}}{{5 \times {{10}^{ – 3}}}} = 3 \times {10^3}N
\end{array}$
Standard 11
Physics
