A particle of mass 'm' moving with velocity ' v' collides inelastically with a stationary particle of mass '2m' . The speed of the system after collision will be

The correct option/answer is: v/3.

Step-by-Step Explanation:

1. Identify the initial conditions of the system before the collision:
- The first particle has a mass of _m and is moving with an initial velocity of _v.
- The second particle has a mass of 2_m and is stationary, meaning its initial velocity is 0.

2. Determine the total initial momentum of the system:
Momentum ($p$) is given by the product of mass ($m$) and velocity ($v$).
The total initial momentum before the collision ($p_{\text{initial}}$) is:

$p_{\text{initial}}=(m\times v)+(2m\times 0)$

Simplifying this:

$p_{\text{initial}}=mv$

3. Analyze the system after the collision:
Since the collision is inelastic and they stick together (implied by "speed of the system after collision"), both particles move together as a single combined mass.
- The total mass of the combined system after the collision is:

$M_{\text{total}}=m+2m=3m$

Let $v^{\prime }$ be the final velocity of this combined system.
The total final momentum of the system ($p_{\text{final}}$) is:

$p_{\text{final}}=3m{v}^{\prime }$

4. Apply the Law of Conservation of Linear Momentum:
In the absence of external forces, the total momentum of the system before the collision is equal to the total momentum of the system after the collision:

$p_{\text{initial}}=p_{\text{final}}$

Substituting the expressions we derived:

$mv=3m{v}^{\prime }$

5. Solve for the final speed $v^{\prime }$:
We can divide both sides by the mass $m$ (assuming $m\ne 0$):

$v=3v^{\prime }$

Rearranging the equation gives:

$v^{\prime }=\frac{v}{3}$

Thus, the speed of the system after the collision will be v/3.