A particle of mass m moving eastward with a speed v collides with another particle of the same mass moving northward with the same speed v . The two particles coalesce on collision. The new particle of mass 2m will move in the north-easterly direction with a velocity

The correct answer is v/√2.

We can solve this problem step-by-step using the law of conservation of linear momentum.

Step 1: Define the coordinate system and initial velocities
Let us represent the eastward direction as the positive x-axis with the unit vector $\hat{i}$, and the northward direction as the positive y-axis with the unit vector $\hat{j}$.
For the first particle of mass $m$ moving eastward with speed $v$, its velocity vector is:
${\overrightarrow{v}}_1=v\hat{i}$

For the second particle of mass $m$ moving northward with speed $v$, its velocity vector is:
${\overrightarrow{v}}_2=v\hat{j}$

Step 2: Calculate the total initial momentum
Before the collision, the total momentum of the two-particle system (${\overrightarrow{P}}_{\text{initial}}$) is the vector sum of their individual momenta:
${\overrightarrow{P}}_{\text{initial}}=m{\overrightarrow{v}}_1+m{\overrightarrow{v}}_2$

Substituting the velocity vectors:
${\overrightarrow{P}}_{\text{initial}}=mv\hat{i}+mv\hat{j}$

Step 3: Calculate the total final momentum
On collision, the two particles coalesce (stick together) to form a single combined particle. The total mass of this new particle is:
$M=m+m=2m$

Let $\overrightarrow{V}$ be the velocity of the coalesced particle after collision. The final momentum (${\overrightarrow{P}}_{\text{final}}$) is:
${\overrightarrow{P}}_{\text{final}}=2m\overrightarrow{V}$

Step 4: Apply the law of conservation of momentum
Since no external forces act on the system, the total momentum is conserved:
${\overrightarrow{P}}_{\text{final}}={\overrightarrow{P}}_{\text{initial}}$

Substituting the momentum expressions:
$2m\overrightarrow{V}=mv\hat{i}+mv\hat{j}$

Divide both sides by $2m$ to find the velocity vector $\overrightarrow{V}$:
$\overrightarrow{V}=\frac{v}{2}\hat{i}+\frac{v}{2}\hat{j}$

Step 5: Calculate the magnitude of the final velocity
The magnitude of the velocity vector is given by:
$V=\sqrt{{\left(\frac{v}{2}\right)}^2+{\left(\frac{v}{2}\right)}^2}$

Simplify the expression inside the square root:
$V=\sqrt{\frac{v^2}{4}+\frac{v^2}{4}}$

$V=\sqrt{\frac{2v^2}{4}}=\sqrt{\frac{v^2}{2}}=\frac{v}{\sqrt{2}}$

Therefore, the new particle of mass 2m will move in the north-easterly direction with a velocity of v/√2.