The velocity of each object after the collision

AI Thread Summary
In an elastic collision involving a 5g object moving at 20cm/s and a stationary 10g object, momentum and energy conservation principles are applied to find the final velocities of both objects. Two equations are established: one for momentum conservation and another for energy conservation. By solving these simultaneous equations, the final velocities can be determined. To find the fraction of energy transferred to the 10g object, the initial kinetic energy is compared to the kinetic energy of the 10g object after the collision. This approach effectively calculates the energy transfer during the collision.
mizzy
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Homework Statement


A 5g object moving to the right with a speed of 20cm/s makes an elastic head on collision with a 10g object initally at rest. Find a) The velocity of each objet after the collision. b) The fraction of the energy transferred to the 10g object.


Homework Equations


In an elastic collision, momentum and energy is conserved.

m1v1 + m2v2 (before collision) = m1v1 + m2v2 (after collision)

v1i - v2i = -(v1f - v2f)


The Attempt at a Solution


where do i start? I tried using the conservation of momentum equation, but you end up with two unknowns, final velocities of both objects.

Can someone guide me please? :confused:
 
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mizzy said:
I tried using the conservation of momentum equation, but you end up with two unknowns, final velocities of both objects.
But you have two equations. Use them both to solve for the final velocities. (The second equation that you list incorporates energy conservation, valid for elastic collisions.)
 
ok, i have my two equations.

first equation :
0.001 = (0.005)V1f + (0.01)V2f

second equation:
0.2 = -V1f + V2f

i might need a refresher is solving simultaneous equations. Can I multiply 0.01 to the second equation?
 
mizzy said:
ok, i have my two equations.

first equation :
0.001 = (0.005)V1f + (0.01)V2f

second equation:
0.2 = -V1f + V2f
Good.
i might need a refresher is solving simultaneous equations. Can I multiply 0.01 to the second equation?
Sure. Then you'd subtract the two equations to eliminate V2f. (You can also multiply the second equation by 0.005 and add the two equations to eliminate V1f.)
 
Thank You!
 
for part b, when it asks for the fraction of energy transferred to the 10g object, is it just the kinetic energy before the collision?
 
mizzy said:
for part b, when it asks for the fraction of energy transferred to the 10g object, is it just the kinetic energy before the collision?
Compare the initial KE before collision to the KE of the 10g object after collision. What fraction of that initial KE went into the 10g object?
 
Oh ok. So what I did was to find the kinetic energy before the collision and also the kinetic energy of the 10g object. Then I took the kinetic energy of the 10g object and divided it by the kinetic energy before the collision.

is that correct?
 
Yes, that's correct.
 

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