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?
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=20081104122100AABK0SOIf 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.)