Question

After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.0 m. Determine the magnitude of the impulse delivered to the ball by the floor.

Answer 1

Given Information:

Mass of ball = m = 0.320 kg

Initial height = h₁ = 19 m

Final height = h₂ = 15 m

Required Information:

Impulse = I = ?

Answer:

Impulse = 11.77 kg.m/s

Explanation:l

We know that impulse is equal to change in momentum

I = Δp

I = p₁ - p₂

I = mv₁ - mv₂

I = m(v₁ - v₂)

Where m is the mass of ball, v₂ is the final velocity of the ball, and v₁ is the initial velocity of the ball.

So first we need to find the initial and final velocities of the ball

The relation between initial potential energy and final kinetic energy before the collision is given by

PE₁ = KE₂

mgh₁ = ½mv₂²

gh₁ = ½v₂²

v₂² = 2gh₁

v₂ = √2gh₁

v₂ = √2*9.8*19

v₂ = 19.3 m/s

The relation between initial kinetic energy and final potential energy after the collision is given by

KE₁ = PE₂

½mv₁² = mgh₂

½v₁² = gh₂

v₁² = 2gh₂

v₁ = √2gh₂

v₁ =√2*9.8*15

v₁ = 17.15 m/s

Finally, we can now find the magnitude of the impulse delivered to the ball by the floor.

I = 0.320(17.5 - (-19.3))

I = 11.77 kg.m/s

Answer 2

Answer:

$Imp = 11.666\,\frac{kg\cdot m}{s}$

Explanation:

Speed experimented by the ball before and after collision are determined by using Principle of Energy Conservation:

Before collision:

$(0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (19\,m) = \frac{1}{2}\cdot (0.320\,kg)\cdot v_{A}^{2}$

$v_{A} \approx 19.304\,\frac{m}{s}$

After collision:

$\frac{1}{2}\cdot (0.320\,kg)\cdot v_{B}^{2} = (0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (15\,m)$

$v_{B} \approx 17.153\,\frac{m}{s}$

The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

$Imp = (0.32\,kg)\cdot [(17.153\,\frac{m}{s} )-(-19.304\,\frac{m}{s} )]$

$Imp = 11.666\,\frac{kg\cdot m}{s}$