Q. A player kicks a football of mass $0.5\ kg$ and the football begins to move with a velocity of $10\ m/s$. If the contact between the leg and the football lasts for $\frac{1}{50}\ sec$, then the force acted on the football should be
A) $2500\ N$
B) $1250\ N$
C) $250\ N$
D) $625\ N$
Explanation
(c) Force on the football $F = m\frac{dv}{dt}$
$F = \frac{m(v_{2} - v_{1})}{dt} = \frac{0.5 \times (10 - 0)}{1/50} = 250N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.