Question

a beryllium nucleus(mass 1.51*10^-26kg) is moving with a speed of 3.0*10^6m/s when it collides a elastically with a stationary deuterium nucleus of mass 3.34*10^-27kg. After the collision both nuclei moves off in the same direction.Determine the velocity of each nucleus after the collision.

Answer 1

I would solve it by using the conservation of energy, but I am not sure that it stands in this context.

Answer 2

I would say that the kinetic energy of this nucleus is
1/2 mv^2
1/2 * 1.51*10^-26 *(3*10^6)^2

Answer 3

and this is equal to
1/2 (1.51*10^-26+3.34*10^-27)(new v)^2

Answer 4

\[m_1V_1 + m_2V_2 (pre collision) = m_1V_1 + m_2V_2 (postcollision)\]
simply insert your numbers and voila

Answer 5

that equation does not seem to be right :-)
you can cross out m1v1
to get
m2v2(pre)=m2v2(post)
v2(pre)=v2(post)

Answer 6

u say 'a elastically' - if its' inelastically' you can solve it by conservation of momentum
- total momentum (mv) is conserved. If its elastic you need to know the coefficient of restitution. since you are not given this i'm guessing its inelastic. I'm with Andras - I'm not sure about conservation of energy being the way to go.

Answer 7

By definition if an collision is elastic, energy is conserved.

For first principles on this topic, watch:
http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-16/

The part you need to answer this question is at the beginning of this lecture.

Answer 8

Oh yeah, than I was right :-)

Answer 9

thanx jamesj - i've learnt something today