Finding two objects velocity after an elastic collision

AI Thread Summary
In an elastic collision involving a 4-kg block moving at 7 m/s and a stationary 3-kg block, the conservation of momentum and kinetic energy equations are applied to find the final velocities of both blocks. The initial momentum equation is set up as 28 = 4(vf1) + 3(vf2). To solve for the final velocities, the relationship between them is established as vf1 = vf2 - 7. Substituting this into the momentum equation yields vf2 = 8 m/s and vf1 = 1 m/s. The discussion emphasizes the importance of recognizing the type of collision to apply the correct principles.
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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