Question:
A ball of 0.1 kg mass is dropped on a hard floor
from a height of 0.45 m and rises to a height
of 0.20 m. If it was in touch with the floor for
0.1 s, the net force it applied on the floor while
bouncing is : (take the gravitational acceleration
g = 10 m s–2)
A ball of 0.1 kg mass is dropped on a hard floor
from a height of 0.45 m and rises to a height
of 0.20 m. If it was in touch with the floor for
0.1 s, the net force it applied on the floor while
bouncing is : (take the gravitational acceleration
g = 10 m s–2)
Options:
- 1.0 N
- 6.0 N
- 3.0 N
- 5.0 N
- 1.0 N
- 6.0 N
- 3.0 N
- 5.0 N
Correct Answer: D) 5.0 N
Explanation:
Speed before bounce u = sqrt(2gh1) = sqrt(2*10*0.45) = 3 m/s.
Speed after bounce v = sqrt(2gh2) = sqrt(2*10*0.20) = 2 m/s.
Force F = m(v – (-u))/t = 0.1 * (2 + 3) / 0.1 = 5 N.
Speed before bounce u = sqrt(2gh1) = sqrt(2*10*0.45) = 3 m/s.
Speed after bounce v = sqrt(2gh2) = sqrt(2*10*0.20) = 2 m/s.
Force F = m(v – (-u))/t = 0.1 * (2 + 3) / 0.1 = 5 N.