A player stops a football weighing $0.5 kg$ which comes flying towards him with a velocity of $10 m/s$ . If the impact lasts for $1 / 50 th$ sec. and the ball bounces back with a velocity of $15 m/s$ , then the average force involved is

250 N

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1250 N

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500 N

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625 N

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Solution

The correct option is D $625 N$
Given, $m = 0.5 k g$
$u = 10 m / s$
$v = - 15 m / s$
$t = \frac{1}{50} s$
As we know that
change in momentum $Δ p = m (v - u$
$Δ p =$ impluse $= F \times t$
$F = \frac{m (v - u)}{t}$
$= \frac{0.5 (- 15 - 10)}{1 / 50}$
$F = \frac{1}{2} \times - 25 \times 50$
$F = - 625 N$

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A player stops a football weighing 0.5 kg which comes flying towards him with a velocity of 10 m/s. If the impact lasts for 1/50 th sec. and the ball bounces back with a velocity of 15 m/s, then the average force involved is

The correct option is D 625 NGiven, m=0.5 kg u=10 m/s v=−15 m/s t=150s As we know that change in momentum Δp=m(v−u Δp=impluse =F×t F=m(v−u)t =0.5(−15−10)1/50 F=12×−25×50 F=−625 N

A player stops a football weighing $0.5 kg$ which comes flying towards him with a velocity of $10 m/s$ . If the impact lasts for $1 / 50 th$ sec. and the ball bounces back with a velocity of $15 m/s$ , then the average force involved is