Question

a fast moving neutron collides elastically and head on with a nucleus of N^14 the percentage of energy lost by the neutron during elastic collision is nearly​

Answer 1

Answer:

Percentage loss of kinetic energy of neutron is 25%

Explanation:

As we know that the collision between neutron and Nitrogen nuclei is perfectly elastic collision

So we will use momentum conservation before and after collision

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

now we know that

$m v_o + 0 = mv_1 + 14m v_2$

$v_o = v_1 + 14v_2$

now by coefficient of restitution equation

$v_2 - v_1 = v_o$

so we have

$v_2 = \frac{2}{15}v_o$

and now we have

$v_1 = -\frac{13}{15}v_o$

Now loss in kinetic energy of the neutron is given as

$\Delta K = \frac{1}{2}mv_o^2 - \frac{1}{2}m(\frac{13}{15}v_o)^2$

$\Delta K = 0.25(\frac{1}{2}mv_o^2)$

percentage loss is given as

$percentage = \frac{\Delta K}{K} \times 100$

$percentage = 25$%

#Learn

Topic : Collision

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