Q 19 Nucleous
Q.19. An unknown nucleus collides with a ${}^{4}He$ nucleus, and after the collision the two nuclei travel in perpendicular directions relative to each other. If kinetic energy is lost in the collision, the unknown nucleus must be
(1) ${}^{28}N$
(2) ${}^{4}He$
(3) ${}^{12}C$
(4) a nucleus with mass lighter than ${}^{4}He$

Dear Student , 

Let a particle A of mass m and initial velocity u makes collision with another particle B of equal mass at rest. Let the collision be elastic. Let v₁ and v₂ be the velocity of two bodies after collision. Let θ be the angle between A and B after the collision.
Using principal of conservation of linear momentum, we have
$$\begin{gathered} mu=m{v}_1+m{v}_2 \\ \Rightarrow u=v_1+v_2 \\ \Rightarrow u^2=\left(v_1+v_2\right).\left(v_1+v_2\right) \\ \Rightarrow u^2=v_1^2+v_2^2+2v_1{v}_2\cos \theta .....\left(1\right) \end{gathered}$$
Further, total kinetic energy before collision = total kinetc energy after collision.
$$\begin{gathered} \frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2 \\ \Rightarrow u^2=v_1^2+v_2^2 .....\left(2\right) \end{gathered}$$
Equating equation (1) and (2), we have
$2v_1{v}_2\cos \theta =0$
For oblique collision, v_1, v₂ is finite. So, cosθ = 0
$$\begin{gathered} \theta ={\cos }^{-1}0 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2}=90° \end{gathered}$$
Thus, particales would move in mutually perpendicular direction after collision.
So option (4) will be correct here 
Regards