1. A steel ball with mass 40.0 g is dropped from a height of 2.00 m onto a horizontal steel slab. The ball rebounds to a height of 1.60 m. Calculate the impulse delivered to the ball during impact.
To calculate the impulse delivered to the ball during impact, we first need to calculate the initial velocity of the ball just before impact with the slab.
Using the formula for gravitational potential energy:
mgh = 1/2 mv^2
Where:
m = mass of the ball = 40.0 g = 0.040 kg
g = acceleration due to gravity = 9.81 m/s^2
h = height = 2.00 m
v = initial velocity
Substitute the values into the equation:
0.040 kg * 9.81 m/s^2 * 2.00 m = 1/2 * 0.040 kg * v^2
0.784 kg m^2/s^2 = 0.020 kg * v^2
v^2 = 0.784 kg m^2/s^2 / 0.020 kg
v^2 = 39.2 m^2/s^2
v = √39.2 m/s
v ≈ 6.26 m/s
Now that we have the initial velocity, we can calculate the final velocity just after impact using the conservation of energy:
mgh = 1/2 mv_f^2
Where:
h = height = 1.60 m
v_f = final velocity
Substitute the values into the equation:
0.040 kg * 9.81 m/s^2 * 1.60 m = 1/2 * 0.040 kg * v_f^2
0.627 kg m^2/s^2 = 0.020 kg * v_f^2
v_f^2 = 0.627 kg m^2/s^2 / 0.020 kg
v_f^2 = 31.35 m^2/s^2
v_f = √31.35 m/s
v_f ≈ 5.60 m/s
Now, we can calculate the change in velocity Δv during the impact:
Δv = v_f - v
Δv = 5.60 m/s - 6.26 m/s
Δv = -0.66 m/s
Finally, we can calculate the impulse delivered to the ball using the formula for impulse:
Impulse = m * Δv
Impulse = 0.040 kg * (-0.66 m/s)
Impulse = -0.0264 kg m/s
Therefore, the impulse delivered to the ball during impact is approximately -0.0264 kg m/s.