Question

A foot ball player kicks a 0.25kg ball and imparts it a velocity of 10m/s. The contact between foot and ball is only for one-fiftieth of a second. The kicking force is?

Answer 1

We know,

$\boxed{\sf{Force=\dfrac{\Delta P}{\Delta t }\longrightarrow \ Newton's \ 2nd \ law}}$

$\sf{\implies F=\dfrac{m\Delta v}{\Delta f}=\dfrac{m(V_f-V_i)}{\Delta f}}$

Given,

• Mass = 0.25 kg

• Final Velocity ($\sf{V_f}$) = 10 m/s

• Initial Velocity ($\sf{V_i}$) = 0 m/s

• Δf = $\sf{\dfrac{1}{50}}$

Applying values to the equation:

$\sf{\implies F=\dfrac{0.25(10-0)}{\dfrac{1}{50}}}$

$\sf{\implies F=0.25\times 10\times 50}$

$\sf{\implies F=125 \ N}$

$\huge{\boxed{\bold{\sf{Force=125 \ N}}}}$