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 )

Step-by-step solution

As per conservation of energy, 1 2 m u^2=m g h aligned u & = 2 g h & = 2 10 0.45 & = 9 =3 ~m / s aligned After rebound some of its kinetic energy gets lost and remaining gets converted into Potential energy Hence, 1 2 m v^2=m g h^ aligned v & = 2 g h^ & = 2 10 0.20 & = 4 =2 ~m / s aligned We know that, F = m v-m u t = m(v-u) t Taking sign consideration F = 0.1(-2-3) 0.1 =-5 ~N . (-ve sign indicated retarding force)