Question

In a nuclear collision, a proton A is incident on a stationary neighbouring proton B with a velocity of 3.00 x 10 m/s. After impact, A moves in the direction BC inclined at 600 with a velocity of 1.50 x 102 m/s. B then moves in the direction BD declined at 0. If BC makes an angle 600with the initial direction (AB) of A and BD makes and angle G with it ; i) Calculate the speed of B along BD and the angle 0
ii) Verify whether or not the collision is elastic.

Answer 1

1) since the angle $G=0$ , the velocity $B$ according to the law of conservation of momentum $m\times v= m\times v_1\times \ cos {600}+ m\times v_2\times \cos{0}$ , where $m$ is the mass of the proton, $v$ is the velocity of proton $A$ before the collision and $v_1$ after, and $v_2$ is the velocity of proton $B$

Hence $v_2=\frac{v-v_1\times \cos {600}}{\cos{0}}=\frac{30-150\times \cos {600}}{\cos{0}}=179.85$ m/s.

2) Since the sum of kinetic energy is conserved for an absolutely elastic shock,

$\frac{m\times v^2}{2}=\frac{m\times v_2^2}{2}+\frac{m\times v_1^2}{2}$

$\frac{ 30^2}{2}=\frac{179.85^2}{2}+\frac{150^2}{2}$

$450{=}\mathllap{/\,}27423.01$

Hence the impact is inelastic.