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Grade: 12th pass
        
A ball of mass 1kg is dropped from hight 9.8 m,strikes with ground and rebounds at hight of 4.9 m ,if the time of contact b/w ball and ground is 0.1 sec then find impulse and avg force acting on ball.
one year ago

Answers : (1)

Arun
24475 Points
							
Dear Aman
 
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
one year ago
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