what is total KE of two pucks before collision
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@Vincent-Lyon.Fr
Just add up the two KE of the pucks.
ok but when should i consider the x and y direction
KE is a scalar quantity, depending on speed, not on velocity. So no need here to bother with x's and y's.
okay so if i want to find the momentum then it is
1/2 mv^2
MV is momentum
Momentum is a vector quantity depending on velocity. So x's and y's have to be considered properly.
okay but if they ask te magnitude of total momentum of the 2 pucks system after collision
the answer is 5J i did not get
the KE is 5J because it is scalar but why the total momentum 5J
Start again with KE and forget momentum right now. 5J is the right answer.
yes
now they are asking total momentum of the pucks
after collision
what will be the kinetic energy after collision ??
Value of total momentum is 5 kg.m/s, it is not 5 J. (different units)
Total momentum before collision = total momentum after collision (always true)
ya i know that but we dont know whether it is elastic or inelastic
for KE to be conserved
Hold on! Can you ask/answer questions one at a time? It's totally confusing.
LOL:)
alright consider the same diagram (2) what is teh total momentum of two puck system after collision
Well, it's the same as BEFORE the collision, so what you have to do is work out total momentum before collision.
got you thx for replying
Since the two momentum vectors are at right angles, their sum can be found using the Pythagorean theorem; the direction can be found using SOH CAH TOA (specifically, the tangent function)
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