A ball of mass 1 kg dropped from 9.8m height , strikes the ground and rebounds to a height of 4.9m .If the time of contact between ball and ground is 0.1s,then find impulse and average force acting on ball.
26ns
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
Answer is wrong
Answer is wrong..the correct answer is 23.89Ns, 238.9N
Answer is wrong..
It's a wrong answer
Correct answer is 23.52Ns,235.2N
The answer given by kundan is almost correct.
But only at the last part where he solve:
9.8*(1+root2)= 9.8*(1+1.414)= 9.8*2.414=23.857
This is where he went wrong
Similarly we can fine force
Thanks for correcting the mistake
Why has -ve sign being used in front of initial velocity ??
What is meaning if square root 2
Actually the correct answer is 23.89Ns and force is 238.9 N but how ? The solution is yet not cleared
......Question asked by Kundan......
Some favourite answers :
(1) 13.85 by Jitin Sharma
~ Wrong
(2) Correction of Jitin by Sireesha
He forgot to do the sum of √2 + 1
~ Okay
~ But How the (9.8 × 2.414) = 23.857
~ Wrong
(3) Chitransh
Last step -----> [9.8 × 2.414] = 23.657 N-s
This all answers are wrong the write answer is 23.52Ns 235.2.
Still if uh multiply 9.8*2.414
It comes out to 23.657 and not 23.857
But the answer given here in my material is 23.857