Conservation of momentum (wrecking ball hits a stationary object)

In summary, the conversation discusses a situation involving a wrecking ball with a mass of .5kg and a velocity of 3.03 m/s colliding with a stationary block with a height of .9 meters and a mass of .06kg. The calculated exit velocity of the ball after the collision is -3.00 m/s, which suggests that the ball bounces back with almost the same speed it had before the collision. However, there is confusion surrounding the conservation of momentum in this scenario due to the involvement of gravity.
  • #1
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Homework Statement
I am struggling with a conservation of momentum problem, that I can't get right. If anyone can help, I would greatly appreciate it. Please note I am a 52-year-old student who is not very good at math.
Relevant Equations
m1v1i + m2v2i = m1v1f + m2v2f
I have a wrecking ball with a mass of .5kg traveling at 3.03 m/s that hits a stationary block .9 meters high, weighing .06kg. I calculated the ball's exit velocity after it hits the block to be -3.00 m/s .

I calculated the final velocity of th block to be 4.2 m/s
Vf = Sqrt 2(g)(h) = sqrt 2(9.8)(.9) = sqrt 17.64 = 4.2

Looking at conservation of momentum where m1v1i + m2v2i = m1v1f + m2v2f

(.5)(3.03) + (.06)(.0) = (.5)(3.0) + (.06)(4.2)

1.51 = 1.5 + .252 (1.77)

the equation is not balanced, so I don't conservation of momentum. What am I doing wrong?
Again any help would be greatly appreciated.4

Thank you
Michael
 
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  • #2
Welcome to PF! It's not clear what you are doing. Can you draw a sketch of the situation? Is this a 1D problem, a 2D problem, is gravity involved?
 
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  • #3
Hi Bob

Thank you for replying so quickly.. Here is a picture (below) of what I am talking about I am trying to calculate the amount of energy converted from one form to another.

The mass of the ball swinging down is .5kg, its velocity is 3.03 m/s . The mass of the block the ball is colliding with is .06kg with a height of .90 m.

I know PE = KE

The potential energy of the swinging ball is is (m)(g)(dh) or (.5)(9.8)(.47) = 2.30 joules
The kinectic energy is the same 1/2 mv^2 or 1/2(.5)(3.03)^2 = 2.30 joules

But my question is conservation of momentum for the collision b/t the ball and the block

I assume this is an inelastic collision

to determine the momentum in the system

m1 = mass of ball m2 =mass of block

The final velocity of the block falling I found by √2(g)(h) or √2(9.8)(.9) = 4.2 m/s

m1vi + m2vi = m1vf + m2vf, the force of momentum should be the same, but when I work out the equation, the forces are unbalanced.

.5(3.03) + .06(0) = .05(3.00) + .06(4.2)

1.51 kg m/s does not equal 1.5 + .252 = 1.77 m/s

sorry for confining people, I am confused myself Michael
1664573447209.png
 
  • #4
SMOKEYWC said:
Hi Bob

Thank you for replying so quickly.. Here is a picture (below) of what I am talking about I am trying to calculate the amount of energy converted from one form to another.

The mass of the ball swinging down is .5kg, its velocity is 3.03 m/s . The mass of the block the ball is colliding with is .06kg with a height of .90 m.

I know PE = KE

The potential energy of the swinging ball is is (m)(g)(dh) or (.5)(9.8)(.47) = 2.30 joules
The kinectic energy is the same 1/2 mv^2 or 1/2(.5)(3.03)^2 = 2.30 joules

But my question is conservation of momentum for the collision b/t the ball and the block

I assume this is an inelastic collision

to determine the momentum in the system

m1 = mass of ball m2 =mass of block

The final velocity of the block falling I found by √2(g)(h) or √2(9.8)(.9) = 4.2 m/s

m1vi + m2vi = m1vf + m2vf, the force of momentum should be the same, but when I work out the equation, the forces are unbalanced.

.5(3.03) + .06(0) = .05(3.00) + .06(4.2)

1.51 kg m/s does not equal 1.5 + .252 = 1.77 m/s

sorry for confining people, I am confused myselfMichael
I can't make much sense of what you are doing. Momentum is not conserved for this whole scenario, as both objects are releasing PE. Momentum is conserved for the collision, which would equate the momentum of the ball immediately before the collison with the sum of the momentum of the ball and block immediately after the collision.

Because gravity is involved, the momentum of the system from start to finish is not conserved.
 
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  • #5
PeroK said:
I can't make much sense of what you are doing. Momentum is not conserved for this whole scenario, as both objects are releasing PE. Momentum is conserved for the collision, which would equate the momentum of the ball immediately before the collison with the sum of the momentum of the ball and block immediately after the collision.

Because gravity is involved, the momentum of the system from start to finish is not conserved.
thank you so much, you helped clarify things a lot.
 
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  • #6
SMOKEYWC said:
I have a wrecking ball with a mass of .5kg traveling at 3.03 m/s that hits a stationary block .9 meters high, weighing .06kg. I calculated the ball's exit velocity after it hits the block to be -3.00 m/s .
It looks like your calculation says that the ball bounces back with pretty much the speed it had before the collision. This would be the case if the ball collided with an immovable wall without energy loss. Obviously your calculation went astray. Perhaps if you told us what question you wish to answer given this situation, we could be of more help.
 
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  • #7
SMOKEYWC said:
I calculated the final velocity of th block to be 4.2 m/s
Vf = Sqrt 2(g)(h) = sqrt 2(9.8)(.9) = sqrt 17.64 = 4.2
The value you calculated would be the speed of an object that fell 0.9 m starting from rest. It doesn't really apply to the problem at hand.

It would help immensely if you could provide us the original problem statement.
 
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  • #8
Also note, momentum is a vector quantity which means direction matters, you cannot mix vertical and horizontal motion as the momentum is separately conserved in the x and y directions when it is conserved.
 
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  • #9
SMOKEYWC said:
Homework Statement:: I am struggling with a conservation of momentum problem, that I can't get right. If anyone can help, I would greatly appreciate it. Please note I am a 52-year-old student who is not very good at math.
What you have written does not qualify as a homework statement.

You have confused us more than the physics is confusing you!
 
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  • #10
Mister T said:
What you have written does not qualify as a homework statement.

You have confused us more than the physics is confusing you!
I disagree with this criticism. We should never blame or shame the poster for asking questions and trying to learn especially when they just joined PF.
 
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  • #11
I thank everyone who has replied. My question was regarding the Rube Goldberg simulation I am required to make for my PHY 101 class. I should have provided much greater detail to my original query. I was extremely frustrated after spending an embarrassingly long time trying to find the errors in my project, and my question was asked through this frustration. Again I apologize for the confusion. I was able to go back and find my mistake and ultimately was able to answer my question.

Again thank you to everyone.
Michael
 
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1. What is the conservation of momentum?

The conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant over time, unless an external force acts upon it.

2. How does a wrecking ball hitting a stationary object demonstrate conservation of momentum?

When a wrecking ball swings and hits a stationary object, the momentum of the wrecking ball is transferred to the object, causing it to move. This demonstrates conservation of momentum because the total momentum of the system (the wrecking ball and the object) remains the same before and after the collision.

3. What factors affect the conservation of momentum in a collision?

The conservation of momentum is affected by the mass and velocity of the objects involved in the collision. The total momentum of the system will be conserved as long as there are no external forces acting on it.

4. Can the conservation of momentum be violated?

No, the conservation of momentum is a fundamental law of physics and cannot be violated. In any closed system, the total momentum will remain constant.

5. How is the conservation of momentum related to Newton's Third Law of Motion?

Newtons's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that the momentum of an object will be transferred to another object in the opposite direction, resulting in the conservation of momentum in the system.

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