Need quick help - Ball swings down, hits block - Speed of the block?

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The discussion revolves around calculating the speed of a block after a ball swings down and hits it. The correct final speed of the block is identified as 0.767 m/s, but the user struggles with their calculations, initially finding incorrect speeds for both the ball and the block. Key concepts discussed include the conversion of potential energy to kinetic energy, with the equation mgh representing potential energy and KE = (1/2)mv² for kinetic energy. The conservation of energy principle is emphasized, stating that the loss of potential energy from the ball equals the gain in kinetic energy of the block. Ultimately, the user is guided to use these energy equations to find the block's speed accurately.
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Homework Statement



Ok I can't seem to figure this out. Maybe I am tired, or strung out on coffee :), but I need some help!

The image shows the question with diagram:

30hqag4.jpg




Homework Equations





The Attempt at a Solution



Iv done this a few different ways, but all WRONG. The CORRECT answer is V = 0.767 m/s

First I found the speed of the ball by going mgh (.300)(9.8)(.500) which gave me 1.47 m/s

Then I found the speed as it left the block after hitting it by the same method. Which would be 0.588 m/s

So then I simply used: m1v1i + m2v2i = m1v1f + m2v2f

(.300)(1.47) + (.200)(0) = (.300)(.588) + (.200)(x) and solve for x gave me: 1.32 m/s

WHAT am I doing WRONG?

Very appreciate any help on this one guys. Thanks!
 
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Personally, I would determine the potential energy that the object has at hi, and then determine the potential energy at hf. The difference between these values will equal the gain in kinetic energy of the box.
 
Can you possibly elaborate on that for me? :)
 
Have you explored any energy theorems in Physics yet, pertaining to potential energy and kinetic energy?
 
sandy.bridge said:
Have you explored any energy theorems in Physics yet, pertaining to potential energy and kinetic energy?

Yes I have.

So, you are saying determine the potential energy at Hi and Potential energy at Hf

Once I have those to peices of data, what do I do with them?
 
You will notice that at h2 the object has a smaller potential energy than it did when it was at hi; that is, energy was transferred from the ball to the block, which can be seen as kinetic energy.
 
so is it like...

So to calculate the potential energy at Height initial: its (m)(g)(h) ?

same with height final? (m)(g)(h) ?
 
Yes, assuming constant mass, we have,
m_{ball}g(h_1-h_2)=\frac{1}{2}mv_{block}^2
 
Is that expression saying that the difference in potential energy = the same amount in Kinetic energy?
 
  • #10
In words, the loss of energy depicted by the swinging ball can be accounted for by the gain in kinetic energy of the box. The potential energy of the ball changed did it not? Law of of conservation of mass implies the energy of the system remains constant, and hence is merely a transfer from the ball to the box.
 
  • #11
So I have then kinetic energy of .882 applied to the block.

How do I use that to find the blocks speed?
 
  • #12
What is the formula for kinetic energy? I already provided it to you.
 
  • #13
nukeman said:
First I found the speed of the ball by going mgh (.300)(9.8)(.500) which gave me 1.47 m/s
mgh yields potential energy in Joules, not speed. You need to use the kinetic energy equation relating velocity to KE in order to find the speed that comes from converting this PE to KE.
 
  • #14
sandy.bridge said:
What is the formula for kinetic energy? I already provided it to you.

KE = 1/2mv^2
 
  • #15
gneill said:
mgh yields potential energy in Joules, not speed. You need to use the kinetic energy equation relating velocity to KE in order to find the speed that comes from converting this PE to KE.

How do you do this? :( Relating Velocity to KE?
 
  • #16
nukeman said:
How do you do this? :( Relating Velocity to KE?

See your post immediately above! KE = (1/2) MV2.

If some potential energy PE = Mgh is converted to kinetic energy KE = (1/2) MV2, then that means KE = PE.

Mgh = (1/2) MV2. Solve for V.
 
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