A rubber ball of mass 18.0 g is dropped from a height of 1.75 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?

You want the change of momentum.

First speed when hitting floor:
(1/2) m v^2 = m g h
v^2 = 2 g h
v^2 = 2*9.8 * 1.75
v = -5.86 (negative because down)

Now initial speed up
v^2 = 2 * 9.8 * 1.55
v = 5.28
change in velocity = 5.28 -(-5.86)
= 11.14
so change in momentum = .018*11.14 = 0.2 kg m/s

The answer is 0,2kgm/s

To find the impulse applied to the ball during the collision, we can use the principle of conservation of mechanical energy.

The initial mechanical energy (Ei) of the ball before the collision is given by the sum of its potential energy (PEi) and its kinetic energy (KEi):

Ei = PEi + KEi

Since the ball is dropped from a height of 1.75 m, the initial potential energy of the ball is:

PEi = m * g * hi

where m is the mass of the ball in kilograms (converted from grams) and g is the acceleration due to gravity (approximately 9.8 m/s^2). hi is the initial height of the ball:

PEi = 0.018 kg * 9.8 m/s^2 * 1.75 m

The final mechanical energy (Ef) of the ball after the collision is given by the sum of its potential energy (PEf) and its kinetic energy (KEf):

Ef = PEf + KEf

Since the ball rebounds to a height of 1.55 m, the final potential energy of the ball is:

PEf = m * g * hf

where hf is the final height of the ball:

PEf = 0.018 kg * 9.8 m/s^2 * 1.55 m

Since the velocity of the ball is reversed by the collision, its final kinetic energy is zero:

KEf = 0

According to the principle of conservation of mechanical energy, the initial mechanical energy is equal to the final mechanical energy:

Ei = Ef

PEi + KEi = PEf + KEf

Simplifying the equation, we have:

PEi - PEf = KEf - KEi

Substituting the values, we can calculate the impulse applied to the ball during the collision.

Impulse = change in momentum

The change in momentum of the ball is given by:

Impulse = m * (vf - vi)

where m is the mass of the ball and vf and vi are the final and initial velocities, respectively.

Since the velocity of the ball is reversed by the collision, the final velocity is negative:

vf = -vi

Therefore, the change in momentum is:

Impulse = m * (-vi - vi) = -2m * vi

Substituting the values, we have:

Impulse = -2 * 0.018 kg * vi

Now, we need to find the initial velocity (vi) of the ball.

Using the equation for potential energy:

PEi = PEf

m * g * hi = m * g * hf

Simplifying the equation, we have:

hi = hf

Solving for vi:

vi = sqrt(2 * g * hi)

Substituting the values, we can calculate the initial velocity of the ball:

vi = sqrt(2 * 9.8 m/s^2 * 1.75 m)

Now that we know the initial velocity (vi) and the mass (m) of the ball, we can calculate the impulse applied to the ball during the collision:

Impulse = -2 * 0.018 kg * vi

To find the impulse applied to the ball during the collision, we can use the principle of conservation of mechanical energy.

The initial potential energy of the ball is given by mgh, where m is the mass of the ball (18.0 g or 0.018 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the initial height (1.75 m). So, the initial potential energy is:

PE_initial = mgh = 0.018 kg × 9.8 m/s² × 1.75 m

The final potential energy of the ball is given by mgh, where h is the final height (1.55 m). So, the final potential energy is:

PE_final = mgh = 0.018 kg × 9.8 m/s² × 1.55 m

According to the conservation of mechanical energy, the difference between the initial and final potential energies is equal to the work done by the impulse:

PE_initial - PE_final = work done

The work done by the impulse is also equal to the product of the impulse and the change in velocity (Δv):

PE_initial - PE_final = impulse × Δv

The change in velocity is equal to the initial velocity (v_initial) minus the final velocity (v_final), and since the velocity is reversed, we have:

Δv = v_initial - (-v_final) = v_initial + v_final

Therefore:

PE_initial - PE_final = impulse × (v_initial + v_final)

Using the equation for potential energy:

PE_initial = mgh

PE_final = mgh

We can substitute these values back into the equation:

mgh - mgh = impulse × (v_initial + v_final)

Simplifying:

0 = impulse × (v_initial + v_final)

Now we can solve for impulse:

impulse = 0 / (v_initial + v_final)

Since the velocity is reversed, the initial velocity (v_initial) is positive and the final velocity (v_final) is negative.

So:

impulse = 0 / (v_initial - v_final)

Now we need to find the velocities.

The initial velocity can be found using the equation:

v_initial = √(2gh)

Plugging in the values:

v_initial = √(2 × 9.8 m/s² × 1.75 m)

Calculating:

v_initial ≈ 5.14 m/s

The final velocity can be found using the equation:

v_final = -√(2gh)

Plugging in the values:

v_final = -√(2 × 9.8 m/s² × 1.55 m)

Calculating:

v_final ≈ -4.79 m/s

Now we can substitute these values back into the equation for impulse:

impulse = 0 / (5.14 m/s - (-4.79 m/s))

Calculating:

impulse ≈ 0 / (5.14 m/s + 4.79 m/s)

impulse ≈ 0 / 9.93 m/s

Therefore, the impulse applied to the ball during the collision is approximately 0 N*s.