A ball B1 of mass M moving northwards with velocity v collides elastically with another ball B2 of same mass but moving eastwards with the same velocity v. Which of the following statements will be true

The correct statement is: B1 moves eastwards and B2 moves north wards.

Let us analyze the collision step-by-step using the principles of conservation of linear momentum and conservation of kinetic energy (since the collision is elastic).

Step 1: Set up the initial conditions using vector notation.
Let the eastward direction be represented by the unit vector $\hat{i}$ (positive x-axis) and the northward direction be represented by the unit vector $\hat{j}$ (positive y-axis).
Both balls have the same mass $M$.
- Ball $B_1$ is moving northwards with velocity $v$, so its initial velocity vector is:
${\overrightarrow{v}}_1=v\hat{j}$
- Ball $B_2$ is moving eastwards with velocity $v$, so its initial velocity vector is:
${\overrightarrow{v}}_2=v\hat{i}$

Step 2: Conservation of Linear Momentum.
In any collision, the total linear momentum is conserved. Let the final velocities of $B_1$ and $B_2$ be ${\overrightarrow{v}}_{1f}$ and ${\overrightarrow{v}}_{2f}$ respectively.
$M{\overrightarrow{v}}_1+M{\overrightarrow{v}}_2=M{\overrightarrow{v}}_{1f}+M{\overrightarrow{v}}_{2f}$
Since the masses are identical, we can divide the entire equation by $M$:
${\overrightarrow{v}}_1+{\overrightarrow{v}}_2={\overrightarrow{v}}_{1f}+{\overrightarrow{v}}_{2f}$
Substituting the initial velocity values:
$v\hat{j}+v\hat{i}={\overrightarrow{v}}_{1f}+{\overrightarrow{v}}_{2f}$ --- (Equation 1)

Step 3: Conservation of Kinetic Energy.
For an elastic collision, total kinetic energy is conserved:
$\frac{1}{2}M{v}_1^2+\frac{1}{2}M{v}_2^2=\frac{1}{2}M{v}_{1f}^2+\frac{1}{2}M{v}_{2f}^2$
Dividing by $\frac{1}{2}M$:
$v^2+v^2=v_{1f}^2+v_{2f}^2$
$2v^2=v_{1f}^2+v_{2f}^2$ --- (Equation 2)

Step 4: Verify the options against these conservation laws.
Let us test the option where B1 moves eastwards (${\overrightarrow{v}}_{1f}=v\hat{i}$) and B2 moves northwards (${\overrightarrow{v}}_{2f}=v\hat{j}$):
- Momentum check:
${\overrightarrow{v}}_{1f}+{\overrightarrow{v}}_{2f}=v\hat{i}+v\hat{j}$
This perfectly matches Equation 1 ($v\hat{j}+v\hat{i}$), satisfying conservation of momentum.
- Kinetic Energy check:
The magnitudes of the final velocities are $v_{1f}=v$ and $v_{2f}=v$.
$v_{1f}^2+v_{2f}^2=v^2+v^2=2v^2$
This perfectly matches Equation 2, satisfying conservation of kinetic energy.

Therefore, the collision results in the two balls simply swapping their velocities along their perpendicular lines of action, meaning B1 moves eastwards with velocity v and B2 moves northwards with velocity v.