Answer to Question #142681 in Mechanics | Relativity for mungaster

Question

In a nuclear collision, a proton A is incident on a stationary
neighboring proton B with a velocity of 3.00 x 10 ms- .
After impact A moves in the direction BC
with a velocity of 1.50 x 10 m/s. B then moves in the direction
BD. If BC makes an angle of 60degrees with the initial direction (AB)
of A and BD makes an angle G with it ;
(i) Calculate the speed of B along BD and the angle teta
verify whether or not the collision is elastic.

Answer 1

Steps to do..

1. Check if velocity of proton is much larger (i.e. closer to speed of light). If velocity is much smaller than that of light then use non-relativistic definitions of energy and momentum.
2. In the above question speed of protons are very small hence using non-relativistic definition of momentum we use conservation of components linear momentum along parallel and perpendicular to initial velocity.
3. Solving the two equation obtained from conservation of momentum we get the speed of proton-B and its angle with line AB.
4. To check if this collision is elastic or not we will check if total initial and final kinetic energies are equal.

Detailed solution...

Speed of light is $c=3\times10^{8}$ m/s. Here proton's speed is very less than "c". So here we will use non-relativistic definitions of energy and momentum. Initial momentum along x-axis is...

$p_{xi}=m\times V_A =m\times 3\times 10=30m.....eq[1]$

final momentum along x-direction...

$p_{xf}= (m\times 1.5\times 10\times \cos{60})+m V_{B}\cos{\theta}\\ p_{xf}=\cfrac{15m}{2}+mV_{B}\cos{\theta}.....eq[2]$

Initial momentum along y-direction...

$p_{yi}=0.....eq[3]$

Final momentum along y-direction...

$p_{yf}= (m\times 1.5\times 10\times \sin{60})-(mV_{b}\sin{\theta}).....eq[4]$

Conservation of momentum...

$p_{xi}=p_{xf}$

and

$p_{yi}=p_{yf}$

From eq[1] and eq[2]...

$30m = \cfrac{15m}{2}+mV_{B}\cos{\theta}\\ \cfrac{45}{2}= V_{B}\cos{\theta}.....eq[5]$

and from eq[3] and eq[4]...

$\cfrac{15m\sqrt{3}}{2}= mV_{B} \sin{\theta}\\ \cfrac{15\sqrt{3}}{2}= V_{B}\sin{\theta}.....eq[6]$

squaring and adding eq[5] and eq[6]...

$\bigg(\cfrac{45}{2}\bigg)^{2}+ \bigg(\cfrac{15\sqrt{3}}{2}\bigg)^{2}= (V_{B})^{2}\\ V_{B}= \cfrac{\sqrt{45^{2}+(15\sqrt{3})^2}}{2}\\ V_{B}=51.96 m/sec.....Ans$

deviding eq[6] by eq[5]...

$\tan{\theta}= \cfrac{1}{\sqrt{3}}\\ \theta= 30\degree.....Ans$

$K.E_{i}= \cfrac{m(3\times10^2)}{2}=450m\\ K.E_{f}=\cfrac{m(1.5\times10)^2}{2}+ \cfrac{mV_{B}^2}{2}\\ K.E_{f}=\cfrac{1}{2}m[15^2+59.96^2]\\ K.E_{f}=1462.5m$

Since $K.E_{i}{=}\mathllap{/\,} K.E_{f}$

hence collision is not elastic.