Qualification > Sciences
physics.....TOUGH QUESTION
nid404:
Could you illustrate that on the graph. I am still not clear. Phy is not too good a subject of mine
zxcvbnm:
someone please explain question 14 of november 09/12...
nid404:
Nov 09/12
Q14) momentum before spring was released=0
so momentum after releasing would also be 0
momentum of 2kg trolley= 4kgm/s
momentum of 1kg trolley would be= -4kgm/s (since sum has to equal to zero)
mv=-4
velocity of 1kg trolley will be 4m/s in the opp direction
total k.e= 1/2 X 2 X 22 + 1/2 X1 X42
=4 + 8
=12J
zxcvbnm:
--- Quote from: nid404 on May 18, 2010, 05:23:20 pm ---Nov 09/12
Q14) momentum before spring was released=0
so momentum after releasing would also be 0
momentum of 2kg trolley= 4kgm/s
momentum of 1kg trolley would be= -4kgm/s (since sum has to equal to zero)
mv=-4
velocity of 1kg trolley will be 4m/s in the opp direction
total k.e= 1/2 X 2 X 22 + 1/2 X1 X42
=4 + 8
=12J
--- End quote ---
thanks so much nid!!!! hey if you have some last minute tips for physics, please tell! oh and the formula for phase difference, (2n+1)lambda/2, what does n stand for??? i've totally forgotten it
nid404:
You're welcome.
Which eqn is this? :S
lol i have no idea
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