Answer to Question #141210 in Mechanics | Relativity for Ada

Question

In a nuclear collision, a proton A is incident on a stationary neighboring 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 an angle G with it ;
ii) Verify whether or not the collision is elastic.

Answer 1

Let a proton $A$ of mass $m$ and velocity $\vec {u_A}$ collides with another stationary proton $B$ .

After collision the proton $A$ moves with velocity $\vec {v_A}$ at an angle $\theta$ with the initial direction AB and the proton $B$ moves with velocity $\vec{v_B}$ at an angle $\phi$ with AB as shown in the figure.

Conservation of linear momentum:

Conservation of linear momentum along x-axis:

$mu_A+m\times 0=mv_A\cos\theta+mv_B\cos\phi \\ \Rightarrow v_B\cos\phi =u_A-v_A\cos\theta \qquad ........(1)$

Conservation of linear momentum along y-axis:

$mv_A\sin\theta=mv_B\sin\phi \\ \Rightarrow v_B\sin\phi=v_A\sin\theta\qquad ...........(2)$

Squaring (1) and (2) and adding them together,

$v^2_B\cos^2\phi+v^2_B\sin^2\phi =(u_A-v_A\cos\theta )^2+v^2_A\sin^2\theta\\ \Rightarrow v^2_B(\cos^2\phi+\sin^2\phi)=(u^2_A-2u_Av_A \cos\theta+v^2_A\cos^2\theta)+v^2_A\sin^2\theta\\ \Rightarrow v^2_B(\cos^2\phi+\sin^2\phi)=u^2_A-2u_Av_A \cos\theta+v^2_A(\cos^2\theta+\sin^2\theta)\\ \Rightarrow v^2_B=u^2_A-2u_Av_A \cos\theta+v^2_A$

where we have used the property $\sin^2x+cos^2x=1$ .

$\therefore v_B=\sqrt{ u^2_A-2u_Av_A \cos\theta+v^2_A}$ ..............(3)

Given $u_A=3.00\times 10^2\ m/s$

$v_A=1.50\times 10^2\ m/s$

$\theta=60\degree$

Substituting these values in equation (3), we get

$v_B=\sqrt{ (3.00\times 10^2)^2-2\times 3.00\times 10^2\times 1.50\times 10^2\times \cos60\degree+(1.50\times 10^2)^2}$

$\Rightarrow v_B=2.56\times 10^2\ m/s$

Thus, the speed of the proton B after collision is $2.56\times 10^2\ m/s$ .

Kinetic energy:

Total kinetic energy before collision = $\frac{1}{2}mu^2_A$

Total kinetic energy after collision $=\frac{1}{2}mv^2_A+\frac{1}{2}mv^2_B$

For the collision to be elastic, the total kinetic energy before collision must be equal to the total kinetic energy after collision i.e, for elastic collision

$\frac{1}{2}mu^2_A=\frac{1}{2}mv^2_A+\frac{1}{2}mv^2_B\\ \Rightarrow u^2_A=v^2_A+v^2_B$

If this condition holds, the collision is elastic. Otherwise, the collision is inelastic.

Let's check it now

$u^2_A=(3.00\times 10^2)^2=9.00\times 10^4$

$v^2_A+v^2_B=(1.50\times 10^2)^2+(2.56\times 10^2)^2=8.80\times 10^4$

Thus we can see that $u^2_A>(v^2_A+v^2_B)$ which implies loss of energy and hence the collision is inelastic.

Answer: (i) The velocity of proton B after collision is $2.56\times 10^2\ m/s$

(ii) The collision is not elastic.