Finding two objects velocity after an elastic collision

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Homework Help Overview

The discussion revolves around a problem involving an elastic collision between two blocks, one moving and the other stationary. The original poster presents the masses and initial velocities of both blocks and seeks to determine their final velocities after the collision.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the conservation of momentum and kinetic energy principles. Some express confusion about the conditions of the collision, questioning whether it is elastic or inelastic. Others suggest that additional information may be necessary to fully resolve the problem.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the problem. Some have provided insights into the relationship between momentum and kinetic energy in elastic collisions, while others have raised concerns about missing information. There is no explicit consensus on the assumptions regarding the collision type.

Contextual Notes

Participants note the potential ambiguity in the problem setup, particularly regarding the nature of the collision (elastic vs. inelastic) and the implications for energy conservation. The original poster expresses uncertainty about how to proceed with the calculations.

DeltaForce
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Homework Statement


A 4-kg block moving at 7m/s makes a head-on collision with a stationary block of mass 3kg. Find the velocities of the two blocks after the collision.
m1=4kg, vi1=7m/s
m2=3kg, vi2 = 0m/s
Trying to find vf1 and vf2

2. Homework Equations

Using the conservation of momentum formula.
m1(vi1)+m2(vi2) = m1(vf1)+m2(vf2)

3. The Attempt at a Solution

Plugging in the numbers I get: 4(7)+3(0) = 4(vf1)+3(vf2)
28 = 4(vf1)+3(vf2)
It is at this point where I'm stuck. How do I find vf1 and vf2? I feel like I'm missing something.
 
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DeltaForce said:

Homework Statement


A 4-kg block moving at 7m/s makes a head-on collision with a stationary block of mass 3kg. Find the velocities of the two blocks after the collision.
m1=4kg, vi1=7m/s
m2=3kg, vi2 = 0m/s
Trying to find vf1 and vf2

2. Homework Equations

Using the conservation of momentum formula.
m1(vi1)+m2(vi2) = m1(vf1)+m2(vf2)

3. The Attempt at a Solution

Plugging in the numbers I get: 4(7)+3(0) = 4(vf1)+3(vf2)
28 = 4(vf1)+3(vf2)
It is at this point where I'm stuck. How do I find vf1 and vf2? I feel like I'm missing something.
What about energy in case of elastic collision?
 
ehild said:
What about energy in case of elastic collision?

Kinetic energy is conserved?
Sorry, I'm a bit lost. Could you elaborate a bit more? Thanks
 
DeltaForce said:
Kinetic energy is conserved?
Yes.
 
DeltaForce said:
Kinetic energy is conserved?
Sorry, I'm a bit lost. Could you elaborate a bit more? Thanks
Just an example:
When a body moving with a uniform velocity v collides with another body at rest, the second body after collision moves with the same velocity as the first one.
This is not possible under normal circumstance as some amount of energy is lost during collision in the form of heat or sound.
 
Pallavi said:
When a body moving with a uniform velocity v collides with another body at rest, the second body after collision moves with the same velocity as the first one.
Only if they have the same mass.
But more generally, if KE is conserved in a straight line collision then v2f-v1f=v1i-v2i, regardless of the masses.
 
On the face of it the information given is insufficient - we don't know whether this is an elastic collision, completely inelastic or somewhere in between ? I would guess elastic as assumed by others above but the OP is correct saying he/she feels there is something missing!
 
neilparker62 said:
we don't know whether this is an elastic collision
The thread title says it is.
 
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haruspex said:
Only if they have the same mass.
But more generally, if KE is conserved in a straight line collision then v2f-v1f=v1i-v2i, regardless of the masses.

Oh ok, I see now. I need to substitute that that equation into the first one.
So 7 = v2f - v1f
v1f = v2f-7
28 = 3(v2f) + 4(v2f-7)
v2f = 8m/s
Thus v1f = 1 m/s
Thank you!
 
  • #10
Pallavi said:
Just an example:
When a body moving with a uniform velocity v collides with another body at rest, the second body after collision moves with the same velocity as the first one.
This is not possible under normal circumstance as some amount of energy is lost during collision in the form of heat or sound.

Thanks for responding!
 
  • #11
DeltaForce said:
Thanks for responding!
Your're most welcome :smile:
 

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