Question

A body of mass m makes an elastic collision with another identical body at rest. Show that if the collision is not head-on the bodies go at right angle to each other after the collision.

Answer 1

According to the question, the collision of the two bodies of mass m is not head-on. Thus, the two bodies move in different directions. 

Let the velocity vectors of the two bodies after collision be v₁ and v₂.

As the collision in the question is elastic, momentum is conserved.

On applying the law of conservation of momentum in X-direction, we get:

\[m u_1  + m \times 0   =   m v_1   \cos  \alpha + m v_2   \cos  \beta       .  .  . (i)\] 

On applying the law of conservation of momentum in Y-direction, we get:

\[0 = m v_1   \sin  \alpha   - m v_2   \sin  \beta           .  .  . (ii)\] 

\[ \Rightarrow m v_1   \sin  \alpha   = m v_2   \sin  \beta\] 

\[\text{ Now }, \] 

\[\left( \frac{1}{2} \right)m u_1^2    +   0   =   \left( \frac{1}{2} \right)m v_1^2    +   \left( \frac{1}{2} \right)m \times  v_2^2 \]  

\[ \Rightarrow    u_1^2    =    v_1^2  +  v_2^2            .  .  . (iii)\] 

\[\text{ On  squaring  equation  (i),   we  get: }\] 

\[ u_1^2    =    v_1^2  \cos^2   \alpha   +    v_2^2    \cos^2 \beta   +   2 v_1  v_2   \cos  \alpha  \cos  \beta\] 

\[\text{ Equating  the  equations  (i)  and  (iii),   we  get }\] 

\[ v_1^2  +  v_2^2    =    v_1^2    \cos^2   \alpha   +    v_2^2    \cos^2 \beta   +   2 v_1  v_2   \cos  \alpha  \cos  \beta\] 

\[ \Rightarrow  {v_1}^2    \sin^2   \alpha   +    {v_2}^2    \sin^2   \beta   =   2 v_1  v_2   \cos  \alpha  \cos  \beta\] 

\[ \Rightarrow   2 {v_1}^2    \sin^2   \alpha   =   2 \times  v_1  \times \frac{v_1 \sin  \alpha}{\sin  \beta}  \cos  \alpha  \cos  \beta\] 

\[ \Rightarrow   \sin  \alpha  \sin  \beta   =   \cos  \alpha  \cos  \beta\] 

\[ \Rightarrow   \cos  \alpha   .   \cos  \beta   -   \sin  \alpha  \cos  \beta   =   0\] 

\[ \Rightarrow   \cos  (\alpha + \beta)   =   0 = \cos  90^\circ\] 

\[ \Rightarrow   \alpha + \beta = 90^\circ\]