A rubber ball of mass 43.5 g is dropped from a height of 2.05 m onto a floor. The velocity of the ball is reversed by the collision with the floor, and the ball rebounds to a height of 1.55 m. What impulse was applied to the ball during the collision?

Question

A rubber ball of mass 43.5 g is dropped from a height of 2.05 m onto a floor. The velocity of the ball is reversed by the collision with the floor, and the ball rebounds to a height of 1.55 m. What impulse was applied to the ball during the collision?

Answer 1

To calculate the impulse applied to the ball during the collision, we can use the principle of conservation of momentum. The impulse is equal to the change in momentum of the ball.

Impulse (J) = Change in momentum (Δp)

Momentum is defined as the product of an object's mass and its velocity. Since the ball was dropped and then rebounded, we need to consider the velocities before and after the collision.

Given:
Mass of the ball (m) = 43.5 g = 0.0435 kg
Initial height (h1) = 2.05 m
Final height (h2) = 1.55 m

Step 1: Calculate the initial velocity (v1) of the ball
The initial velocity is not given directly, but we can use the principle of conservation of energy. At the initial height, the potential energy of the ball is converted into kinetic energy as it falls.

Potential Energy (PE) = Mass × Gravitational Acceleration × Height
Kinetic Energy (KE) = 0.5 × Mass × Velocity^2

At the initial height:
PE1 = KE1
m × g × h1 = 0.5 × m × v1^2

Solve for v1:
v1 = √(2 × g × h1)

Here, g is the acceleration due to gravity, which is approximately 9.8 m/s².

Step 2: Calculate the final velocity (v2) of the ball
Similar to step 1, we can use the principle of conservation of energy to calculate the final velocity of the ball as it rebounds.

At the final height:
PE2 = KE2
m × g × h2 = 0.5 × m × v2^2

Solve for v2:
v2 = √(2 × g × h2)

Step 3: Calculate the change in momentum (Δp)
The change in momentum is the difference between the initial and final momenta.

Change in momentum = Final momentum - Initial momentum
Δp = (m × v2) - (m × v1)

Step 4: Calculate the impulse (J)
The impulse is the product of the change in momentum and the mass.

Impulse (J) = Δp × m

Let's substitute the values and perform the calculations:

v1 = √(2 × 9.8 × 2.05) ≈ 6.31 m/s
v2 = √(2 × 9.8 × 1.55) ≈ 5.37 m/s

Δp = (0.0435 kg × 5.37 m/s) - (0.0435 kg × 6.31 m/s)
Δp ≈ -0.0381 kg·m/s (Note: The negative sign indicates a reversal in velocity)

J = Δp × m = -0.0381 kg·m/s × 0.0435 kg
J ≈ -0.00165 kg·m/s

The impulse applied to the ball during the collision is approximately -0.00165 kg·m/s.

A rubber ball of mass 43.5 g is dropped from a height of 2.05 m onto a floor. The velocity of the ball is reversed by the collision with the floor, and the ball rebounds to a height of 1.55 m. What impulse was applied to the ball during the collision?

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To calculate the impulse applied to the ball during the collision, we can use the principle of conservation of momentum. The impulse is equal to the change in momentum of the ball.