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 can use the impulse-momentum theorem, which states that the impulse delivered to an object is equal to the change in its momentum.
First, we need to calculate the initial velocity of the ball just before impact using the formula for final velocity when an object is dropped from a certain height:
v^2 = u^2 + 2as
where
v = final velocity (0 m/s since the ball momentarily comes to a stop at the peak of the bounce)
u = initial velocity (unknown)
a = acceleration due to gravity (9.81 m/s^2)
s = distance fallen (2.00 m)
Rearranging the equation gives us:
u = sqrt(v^2 - 2as)
u = sqrt(0 - 2(9.81)(-2.00))
u = sqrt(39.24)
u = 6.27 m/s
Next, we calculate the change in momentum of the ball:
Δp = mvf - mvi
Δp = 0 - (0.04 kg)(6.27 m/s)
Δp = -0.251 kg·m/s
Since impulse is equal to the change in momentum:
Impulse = -0.251 kg·m/s
Therefore, the impulse delivered to the ball during impact is -0.251 kg·m/s.