A player stops a football weighing 0-5kg which comes flying towards him with a velocity of 10m-s- If the impact lasts for 150th sec and the ball bounces back with a velocity of 15m-s- then the average force involved is -

The correct option is D 625N Initial momentum = (0.5 × 10) î Final momentum = (0.5 × 15) î Change in momentum = 7.5 î 5 î #160; #16 ...

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Standard XII

Physics

Work Energy Theorem

A player stop...

Question

A player stops a football weighing $0.5 k g$ which comes flying towards him with a velocity of $10 m / s$ . If the impact lasts for $\frac{1}{50} t h$ 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 $625 N$
Initial momentum = $(0.5 \times 10)^i$
Final momentum = $- (0.5 \times 15)^i$
Change in momentum = $- 7.5^i - 5^i$
= $- 12.5^i$
Force involved = $\frac{12.5}{(\frac{1}{50})} = 625 N$

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A player stops a football weighing 0.5kg which comes flying towards him with a velocity of 10m/s. If the impact lasts for 150th sec and the ball bounces back with a velocity of 15m/s, then the average force involved is :

The correct option is D 625NInitial momentum = (0.5×10)^iFinal momentum = −(0.5×15)^iChange in momentum =−7.5^i−5^i = −12.5^i Force involved = 12.5(150)=625N

A player stops a football weighing $0.5 k g$ which comes flying towards him with a velocity of $10 m / s$ . If the impact lasts for $\frac{1}{50} t h$ sec and the ball bounces back with a velocity of $15 m / s$ , then the average force involved is :

The correct option is D $625 N$
Initial momentum = $(0.5 \times 10)^i$
Final momentum = $- (0.5 \times 15)^i$
Change in momentum = $- 7.5^i - 5^i$
= $- 12.5^i$
Force involved = $\frac{12.5}{(\frac{1}{50})} = 625 N$