Answer:
The impulse received by the ball from the floor during the bounce is approximately 1.11329438 m·kg/s
Explanation:
The given mass of the ball, m = 86 g = 0.089 kg
The height from which the ball is dropped, H = 2.87 m
The height to which the ball bounces, h = 1.28 m
Mathematically, we have;
Δp = F·Δt
Where;
Δp = The change in momentum = m·Δv
F = The applied force
Δt = The time of contact with the force
The velocity of the ball just before it touches the ground, v₁ = -√(2·g·H)
The velocity with which the ball leaves, v₂ = √(2·g·h)
The change in momentum, Δp = m·(v₂ - v₁)
∴ Δp = m·(√(2·g·h) - (-√(2·g·H))) = m·(√(2·g·h) +√(2·g·H) )
The impulse, Δp, received by the ball from the floor during the bounce is given as follows;
Δp = 0.089 kg × (√(2 × 9.8 m/s² × 1.28 m) + √(2 × 9.8 m/s² × 2.87 m)) ≈ 1.11329438 m·kg/s
The impulse received by the ball from the floor during the bounce, Δp ≈ 1.11329438 m·kg/s
, after a perfectly elastic collision, the object A is supposed to move with the same velocity in the opposite direction.