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ruby92:
a nucleus of mass 1.67*10^-27 kg moving with a velocity of 2*10^4 collides head on with a boron nucleus of mass 1.7*10^-27kg orginally at rest.
find the velocity of the boron nucleus if the collison is perfectly elastic?

A.T:

--- Quote from: ruby92 on March 25, 2010, 10:38:57 am ---a nucleus of mass 1.67*10^-27 kg moving with a velocity of 2*10^4 collides head on with a boron nucleus of mass 1.7*10^-27kg orginally at rest.
find the velocity of the boron nucleus if the collison is perfectly elastic?

--- End quote ---
Data you know:
m1=1.67*10^-27 kg
m2=1.7*10^-27 kg
u1=2*10^4 ms-1
u2=0
In elastic the total momentum(mv) and the energy(1/2 mv^2) is conserved.
For conserved momentum u have eq:(denote as eq 1 )
m1u1+ m2u2 =  m1v1 + m2v2
And for energy conserved the eq is: (denote as eq 2)
(1/2) m1 u1² + (1/2) m2 u2² = (1/2) m1 v1² + (1/2) m2 v2²

now substitute given values in both eq (1 and 2) and solve them simultaneously and u'll get values of v1 and v2. v2 will be the velocity of the boron nucleus.
this link can help u for your question: http://answers.yahoo.com/question/index?qid=20081104122100AABK0SO

If collisions were INelastic then:(Just for ur concept)
m1=1.67*10^-27 kg
m2=1.7*10^-27 kg
u1=2*10^4 ms-1
u2=0
Using m1u1+ m2u2 =  m1v1 + m2v2
as the collision is perfectly elastic the final velocity of both objects after collision would be zero,and knowing that u2=0 we get:
m1u1=V(m1+m2)
put the given values and you will get v = 9910 ms-1 (correct to 3 significant fig.)

ruby92:
ok a mass of 8kg moving at 6ms-1 towards another block of mass 4kg at 3ms-1
its a head on collison
how would u find the momentum transferred from one block to the other. if its elastic? and if its inelastic?

tnkx  :)

astarmathsandphysics:
For inelastic you would have to know the velocity of one arter the collision then do m(v-u) for inelastic use restitution principle.

A.T:

--- Quote from: ruby92 on March 25, 2010, 01:02:50 pm ---ok a mass of 8kg moving at 6ms-1 towards another block of mass 4kg at 3ms-1
its a head on collison
how would u find the momentum transferred from one block to the other. if its elastic? and if its inelastic?

tnkx  :)

--- End quote ---
If inelastic: (they would become one particle(join) after collision and will have the same velocity)
formula for inelastic:
m1u1 + m2u2 = V(m1+m2)
Use it to find V(the combined velocity) multiply it by masses to get the momentum.

For elastic u'll have to do all the long working shown in last last question,but as far as i know in A-levels they give u a value of one of the final velocity which makes it easier and u can simply find it using
m1u1+ m2u2 =  m1v1 + m2v2

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