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)
1.0 N
6.0 N
3.0 N
5.0 N
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC NDA-1 – 2024
First, calculate the velocity of the ball just before impact (v₁) and just after impact (v₂). Using energy conservation (v² = 2gh), v₁ = sqrt(2 * 10 * 0.45) = 3 m/s (downward) and v₂ = sqrt(2 * 10 * 0.20) = 2 m/s (upward). Let’s take upward as positive. v₁ = -3 m/s, v₂ = +2 m/s. The change in momentum of the ball is Δp = m(v₂ – v₁) = 0.1 kg * (2 – (-3)) m/s = 0.1 * 5 = 0.5 kg m/s (upward). The average net force on the ball during contact is F_net_on_ball = Δp / Δt = 0.5 Ns / 0.1 s = 5 N (upward). This net force is the sum of the upward normal force (N) from the floor and the downward gravitational force (mg): F_net_on_ball = N – mg. So, 5 N = N – (0.1 kg * 10 m/s²) = N – 1 N. This gives the average normal force from the floor on the ball, N = 6 N (upward). By Newton’s third law, the force applied by the ball on the floor is equal in magnitude and opposite in direction to the normal force from the floor on the ball. Thus, the force applied on the floor is 6 N (downward). The question asks for the magnitude of this force.