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A ball ok mass 1kg is dropped from height 9.8m strikes with ground and rebounds at height ok 4.9m if the time ok contact between ball and ground is 0.1 sec than find impuse and average force acting on ball

warish alam , 7 Years ago
Grade 12th
anser 1 Answers
Arun

Last Activity: 7 Years ago

Dear student
 
Velocity when it hits the ground 
u = sqrt(2gH) 
= sqrt(2 * 9.8 * 9.8) 
= 9.8 sqrt(2) m/s 

Velocity just after hitting the ground 
v = sqrt(2gH) 
= sqrt(2 * 9.8 * 4.9) 
= 9.8 m/s 

Impulse = change in momentum 
= mass × change in velocity 
= m * (v - u) 
= 1kg * (9.8 - (-9.8*sqrt(2)) m/s 
= 9.8 * (1 + sqrt(2)) kgm/s 
= 9.8 * 1.414 kgm/s 
= 13.8572 kgm/s 

Average Force = Impulse / Time taken 
= 13.8572 / 0.1 
= 1.38 N
 
Regards
Arun(askIITians forum expert)

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