Calculating Final Velocity of Mass 1 in an Elastic Collision

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In the discussion about calculating the final velocity of mass 1 in an elastic collision, participants analyze a scenario involving two masses, m1 and m2, with given speeds and angles. The key focus is on applying the conservation of momentum and correctly breaking down the momentum into x and y components. Participants struggle with the calculations and the appropriate use of equations for elastic collisions, particularly in determining the final velocity of m1 after the collision. The conversation emphasizes the importance of vector components in momentum and kinetic energy calculations, leading to a resolution of the kinetic energy lost during the collision. Ultimately, the correct approach involves separating the momentum components and applying the conservation laws accurately.
  • #31
ScienceGeek24 said:
p=mv

So you have p, and you have m...
 
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  • #32
hmmm still not 3.59m/s i divided the momentum of px/m=v and it gave me 1.79m/s ...
 
  • #33
ScienceGeek24 said:
But this is an ellastic collision wouldn't it be vf=(2m/m1+m2)vx and vf=(m1-m2/m1+m2)vy ??

What would vx and vy be in that formula? Both objects are initially moving. That formula might apply if one object was moving and the other stationary.

Conservation of momentum always works.
 
  • #34
ScienceGeek24 said:
hmmm still not 3.59m/s i divided the momentum of px/m=v and it gave me 1.79m/s ...

The velocity, like momentum, has components. Divide both of the momentum components by m1 to give the components of the velocity. Then, to find the speed, find the magnitude of that vector.
 
Last edited:
  • #35
Got it! Thanks man thankyou for your patience.
 
  • #36
ScienceGeek24 said:
Got it! Thanks man thankyou for your patience.

No problem. Glad to help. Good luck!
 
  • #37
Wait! on emore thing! I'm trying to find the kinetic enrgy lost i this one and this is what i did KEf= 1/2(8kg+7kg)(3.52)^2=95.052 j and Ke total= 1/2(8kg)(5m/s)^2+1/2(7kg)(3m/s)^2=131.5 J than i subtracted Ktotal -KEf= 44... something something and that is not he right answer the right asnwer is 86.4J . What did i do wrong??
 
  • #38
sciencegeek24 said:
wait! On emore thing! I'm trying to find the kinetic enrgy lost i this one and this is what i did kef= 1/2(8kg+7kg)(3.52)^2=95.052 j and ke total= 1/2(8kg)(5m/s)^2+1/2(7kg)(3m/s)^2=131.5 j than i subtracted ktotal -kef= 44... Something something and that is not he right answer the right asnwer is 86.4j . What did i do wrong??

Only mass m1 is moving after the collision. Don't sum the masses, they are not stuck together.

KE1 = 131.5 j
KE2 = 45.1 j

KE1 - KE2 = 131 - 45.1 = 86.4 j
 
  • #39
which mas did you get the 45.1 with? because i did 1/2(7kg)(3.56)^2=44.35 not 45.1.

Sorry to take all your time man... :P
 
  • #40
opps never mind got it! Thanks gain man!
 

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