Question

A cricket ball of mass 90 g moving with a velocity of 0.5 m/s is stopped by a player
in 0.45 s. What is the force applied by the player to stop the ball?​

Answer 1

Step-by-step explanation:

F=ma

= 90*0-0.5/0.45

=90*0.5/0.45

=90*0.9

=81 N

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Answer 2

$\sf \large \fcolorbox{r}{pink}{ \huge{Solution :)}}$

Given ,

• Mass of ball (M) = 90 g or 0.09 kg
• Initial velocity of ball (U) = 0 m/s
• Final Velocity of ball (V) = 0.5 m/s
• Time taken to stop a ball (T) = 0.45 s

We know that ,

$\large \sf \fbox{F = MA}$

It can be written as ,

$\large \sf \fbox{F = M \times (\frac{V - U}{T} )}$

Substitute the known values , we get

$\mapsto \sf F = 0.09 \times( \frac{0.5 - 0 }{0.45}) \\ \\ \mapsto \sf F = 0.09 \times \frac{50}{45} \\ \\ \mapsto \sf F = 0.09 \times 1.11 \\ \\ \mapsto \sf F = 0.0999 \: \: newton \: \: \{ approx\}$

Hence , 0.0999 newton force applied by the player to stop the ball

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