momentum physics?

Question

After dropping from a height 1.50 m onto a
concrete floor, a 50.0-g ball rebounds to a
height of 0.90 m.
a. Find the impulse acting on the ball as it
dropped.
b. Find the impulse acting on the ball on its
rebound.
c. Find the impulse on the ball while it was
in contact with the floor.

Answer 1

The impulse experienced by an object is the integral with respect to time of the force on that object:

I = integral of {F dt}

For a constant force, I = F * delta-t (i.e., the product of the force and the time over which the force acts.)

Alternatively, because F = dp/dt, where p is the momentum:

I = integral {dp/dt dt} = integral {dp} = delta-p.

That is, the impulse is equal to the change in momentum.

Note that the impulse, like force and momentum, is a vector quantity.


The change in kinetic energy of the ball as it falls is equal to:

m*g*h = = 0.5*m*(delta-v)^2
2*g*h = (delta-v)^2
delta-v = sqrt(2*g*h)

In this case, delta-h = 1.5 m, so

delta-v = sqrt(2 * 9.81 m/s^2 * 1.5 m) = 5.424 m/s

This means that the impulse during the time the ball falls ( = the change in momentum during this time) is:

I = m*delta-v = 0.05 kg * 5.424 m/s = 0.271 kg*m/s = 0.271 N*s


Using the same logic to calculate the impulse as the ball rebounds, the change in kinetic energy is:

m*g*h = 0.5 * m * (delta-v)^2

delta-v = sqrt(2*g*h) = sqrt(2 * -9.81 m/s^2 * 0.9 m) = -4.201 m/s (where this time, the negative root is appropriate, because the ball is slowing down as it rebounds)

I = delta-p = m*delta-v = 0.05 kg * -4.201 m/s = -0.21 N*s

Now, assuming that the ball was at rest when it started falling from its initial position 1.5 m above the ground, we know that the ball's velocity just before it hits the ground was 5.424 m/s and its velocity as it just leaves the ground as it starts to rebound is -4.201 m/s. The impulse while the ball was in contact with the floor was:

delta-p = I = 0.5 kg *(-4.201 m/s - 5.424 m/s) = -0.481 N*s

(The previous answerer forgot that velocity is a vector, and that during contact with the ground, the ball's velocity changes direction. His/her answer to the last part of this question is not correct.)

Answer 2

Hi Kyle B

First let's define impulse. Impulse is an instantaneous change in momentum: dp. There are two ways we can work out dp:
i) dp = mdv (m mass of an object, dv change in velocity)
ii) dp = int|F.dt, or F*t for constant force F.

To answer your questions we'll also need the equations of motion for an object undergoing uniform acceleration.

Let's approach question (a) via both methods to be sure we're right:
a.i) The ball falls from height 1.5m. We use equation
v^2 = u^2 + 2as
to determine the change in velocity dv, and dp = mdv. The initial velocity, u, = 0. a = g, and s = 1.5m Plug in the values and you get
dv=5.4m/s
dp = 0.27kgm/s

a.ii) The ball falls under constant force F=mg for time period t. We use equation
s = ut + 1/2*at^2
to determine t, and dp = F*t. u = 0, s=1.5m, a=g.
t = 0.55s
dp = 0.27kgm/s

Our answers agree. :o)

b) for this question we'll use dp = mdv, and again use v^2 = u^2 + 2as to determine dv. Note that this time v = 0 because the ball will reach a stationary point at the top of its bounce. The initial velocity u is the unknown, and dv = u - v, so dv = u (because v=0):
u^2 = -2gs
u = dv = 4.2m/s
dp = mdv = 0.21kgm/s

--> as an exercise, use the relation dp = F*t to confirm that the answer for dp above is right.


c) In question (a.i) we determined that the final velocity of the ball when it reached the floor was 5.4m/s. In question (b) we determined that the initial velocity of the ball as it left the floor was 4.2m/s. This means that as the ball bounced it underwent a velocity change (dv) of:
dv = v - u
dv = 5.4 - -4.2
dv = 9.6m/s
Therefore the impulse (change in momentum), dp, is:
dp = mdv
dp = 0.05kg*9.6m/s
dp = 0.48kgm/s

(thanks to hfshaw for pointing out my missing minus sign. :o) )


Hope this helps!
The Chicken