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

\[elasticity =0\] because both nuclei moves off in the same direction so they moves with the same velocity after collides.
\[mA \times VA + mB \times VB = (mA+mB) V'\]

\[(1.5 \times10^{-26}) (3 \times 10^{6}) + (3.34 \times 10^{-2}) (0) = (1.5 \times10^{-26}+3.34 \times 10^{-2}) V'\]

Answer 2

ups sorry get bad connection

Answer 3

\[(4.5 \times 10^{-20}) = (1.5\times10^{-26} + 3.34 \times10 ^{-2}) V'\]
\[V' = (4.5 \times10 ^{-20}) / (1.5 \times10^{-26} + 3.34 \times 10^{-2})\]

Answer 4

No, that's wrong. They do NOT move off with the same velocity. Watch the beginning of this lecture to get the equations here straight:

http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-16/

Answer 5

im sorry james J i couldnt see your link cause i get bad connectoin here

Answer 6

Momentum is conserved:
\[m_1v_1 + m_2v_2 = m_1v_3 + v_2v_4\]
Elastic collision => kinetic energy is conserved:\[\frac{1}{2}m_1v_1\text{}^2+\frac{1}{2}m_2v_2\text{}^2=\frac{1}{2}m_1v_3\text{}^2+\frac{1}{2}m_2v_4\text{}^2\]

Answer 7

Given your known quantities (m1, m2, v1, v2), you will end up with two equations with v3, and v4 as unknowns.

Answer 8

We must remember this, that in an elastic collision momentum is conserved. That is the total momentum before collision is equal to the total momentum after. Hence we can express this mathematically as\[m_1u_1+m_2u_2=m_1v_1+m_2v_2\]where m is the mass of the particles, and u and v are the velocities before and after collision respectively, and subscripts identify each particle. So plug in the numbers \[(1.51\times10^{-26})(3\times10^6)+(3.34*10^{-27})(0)=m_1v_1+m_2v_2\]. The problem here is that we have 1 equation, and two unknowns (i.e. we need to know either \(v_1\) or \(v_2\)). We can get around this by assuming that the two particles stick together, and hence they will move off with the same velocity, but a combined mass. the above equation then becomes \[(1.51\times10^{-26})(3\times10^6)+(3.34*10^{-27})(0)=(m_1+m_2)v\]and hence we can obtain teh answer of \[v=\frac{(1.51\times10^{-26})(3\times10^6)}{+(3.34*10^{-27})+(1.51\times10^{-26})}=\frac{4.53\times10^{-20}}{1.844\times10^{-26}}=2.457\times10^6\rm{m/s}\]

Answer 9

So noras would be correct in his argument

Answer 10

No, it is not true that you must assume they stick together and frankly it violates the assumption that the collision is elastic. Seriously, this is first year physics and you if you're answering this question you should know it.

Watch the first six minutes of the MIT lecture I posted and you'll get exactly the equations for v1 and v2 after the collision.

http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-16/

Answer 11

No it is not true to make that assumption. Noras (and mine) is correct if you use the assumption but as you state, this does violate the whole premise of it being an elastic collision (since by them sticking together you loose kinetic energy through internal friction if I remember correctly). I'm being an idiot and forgetting that you can solve for the unknowns by using the conservation of energy, and indeed you must for an elastic collision.

Thanks for reminding me. It has been many years since I have done such a problem.