Question

a ball of 2kg drops vertically onto the floor with a velocity of 20m/s.It rebouds with an initial velocity of 10m/s , impulse acting on the ball during contact will be

Answer 1

Answer:

20 kg-m/s

Explanation:

Impulse = Force×Time

= m×a×t ______(1)

Now We have:

m= 2kg

v= 20m/s

u= 10m/s

a= 9.8m/s^2

Applying eqn: v=u+at, we get:

20 =10+9.8×t

=> t= 10/9.8 s

Applying value of (t) in eqn (1) we get:

Impulse = 2×9.8×10/9.8 = 20kg-m/s