Posted by nammilee_2012 on Sunday, November 25, 2012 at 3:05am.
A 0.500 kg ice puck, moving east with a speed of 3.30 m/s, has a headon collision with a 0.900 kg puck initially at rest. Assuming a perfectly elastic collision, (a) what will be the speed of the 0.500 kg object after the collision? (b) What will be the speed of the 0.900 kg object after the collision?

Science  nammilee_2012, Sunday, November 25, 2012 at 3:06am
Suppose,
The velocity of .5 kg ice puck before collision = u1= 3.30 m/s
The velocity of .9 kg ice puck before collision = u2= 0
The velocity of .5 kg ice puck after collision = v1=?
The velocity of .9 kg ice puck after collision = v2=?
According to conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
=> (u1v1)m1 = (v2u2)m2 ....(i)
Again, conservation of energy gives
1/2 m1u1^2+ 1/2 m2u2^2 = 1/2 m1v1^2 + 1/2 m2v2^2
=> m1(u1^2  v1^2) = m2(v2^2  u2^2) .....(ii)
Divide (ii) by (i)
(u1^2  v1^2)/(u1v1) = (v2^2  u2^2)/(v2u2)
=> u1+v1 = v2+u2
=> v1= u2+v2u1 = v2u1 ....(iii). [u2=0]
Now, substitute v1 in (i)
(u1v2+u1)m1 = (v2u2)m2
=> 2m1u1  m1v2 = m2v2
=> (m1+m2)v2 = 2m1u1
=> v2 = (2m1u1)/(m1+m2) ....(iv)
Substitute v2 in (iii)
v1= v2u1 = (2m1u1)/(m1+m2) u1
=> v1 = (2*.5*3.30)/(.5+.9)  3.30 = 2.357  3.30 = 0.943 m/s
() sign indicates that the ice puck is moving to the west.
Now substitute v1 in (iii) again
v1 = v2u1
=> v2 = u1+v1 = 3.30  0.943 = 2.36 m/s
It means it is going to the east.

Science  Elena, Sunday, November 25, 2012 at 4:59am
Correct answers.
http://hyperphysics.phyastr.gsu.edu/hbase/elacol2.html#c3
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