Impulse steel ball falling on steel slab

In summary: So now, we havev(1)^2 = v(f)^2 - 2gyv(1)^2 = 0 - (-19.6)v(1) = -sqrt(19.6) = -4.43 m/sP(1) = mv(1) = (-.176 kg*m/s)and for step two,v(f)^2 = v(1)^2 - 2gyv(f)^2 = 19.6 - 2(-9.8)(1.60) = 35.2v(f) = 5.93 m/s
  • #1
offbeatjumi
28
0

Homework Statement



mass of steel ball = 40g = .04 kg
h1 = 2.00 m
h2 = 1.60 m
A steel ball is dropped from height of 2.00 m onto horizontal steel slab, rebounds to height of 1.60m. Calculate impulse of ball during impact.


Homework Equations



impulse J = F*dt
J = P2 - P1
I tried using


The Attempt at a Solution



I tried using equations of kinematics to solve for the velocities of the ball before and after impact, to then find the final and initial momentum, and solve for impulse by finding change in momentum.
I have the answer from the back of the book, 0.47 N*s. But I did not get that answer with my method. How should I approach this? Thanks.
 
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  • #2
Your method is correct. It is probably its implementation that is incorrect. Maybe you missed a minus sign, maybe you punched in the wrong numbers. If you show the details of your calculation, someone might be able to locate where you went wrong.
 
  • #3
I divided it in two steps, step one before impact and step two after impact.

in step one:

I use v(f)^2 = v(i)^2 - 2gy
v(f)^2 = 0 - (-39.2)
v(f) = -sqrt(39.2) = v(1)
P(1) = mv(1) = -.2504 kg*m/s

in step two:
I use the same kinematic equation
v(f)^2 = v(1)^2-2gy
v(f)^2 = 39.2 - 2*(-9.8)*1.60 = 70.56
v(f) = 8.4 m/s = v(2)
P(2) = mv(2) = 0.336

P(2) - P(1) = 0.336 - (-.2504) = 0.5863 N*m

answer should be 0.47 N*m
??
 
  • #4
The complete kinematic equation is

[tex]v^{2}_{f}=v^{2}_{i}+2a(x_{f}-x_{i})[/tex]

Here vf is the speed when the object is at position xf and vi is the speed when the object is at xi.

We assume that position is measured from the slab.

For step 1, xi=2.00 m, xf=0, vi=0, a = -9.8 m/s2 and we are looking for vf, the speed of the ball just before it hits the slab on its way down. This gives

[tex]v^{2}_{f}=0^{2}+2(-9.8\;m/s^2)(0 - 2.00\;m)[/tex]

If you do the calculation, you get -.2504 kg*m/s for the momentum as you have already found.

The problem with your solution is the identification of the variables after the bounce. After the bounce the speed of the ball at xi = 0 is the unknown speed vi. What is the speed vf when the position is xf = 1.60 m?
 
  • #5
thank you very much

yeah i did not assign the correct positions/velocities of the ball to the equation, i see now that the final velocity in the second situation is zero
 

1. What is impulse and how does it relate to the steel ball falling on the steel slab?

Impulse is a measure of the change in momentum of an object. In this scenario, impulse is the force and time required for the steel ball to come to a stop after hitting the steel slab. It is directly related to the impact force and the duration of the impact.

2. How is the impulse of the steel ball determined?

The impulse of the steel ball can be determined by multiplying the impact force by the duration of the impact. This can be calculated using the formula: impulse = force x time.

3. What factors can affect the impulse of the steel ball?

The impulse of the steel ball can be affected by factors such as the mass and velocity of the ball, the material and surface of the steel slab, and the angle and height from which the ball is dropped.

4. What is the significance of studying the impulse of the steel ball falling on the steel slab?

Studying the impulse of the steel ball falling on the steel slab can provide valuable insights into the principles of Newton's laws of motion and the conservation of momentum. It can also have practical applications in understanding and predicting the behavior of objects in collisions and impact events.

5. Can the impulse of the steel ball be changed?

Yes, the impulse of the steel ball can be changed by altering any of the factors that affect it, such as the mass, velocity, and surface properties of the ball and the slab. This can also be achieved by changing the angle and height from which the ball is dropped, or by introducing external forces during the impact.

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