A particle of mass 1g having velocity 3î -2ĵ has a glued impact with another particle of mass 2g and velocity as 4ĵ - 6k̂. Velocity of the formed particle is

The correct option is 4.6 ms⁻¹.

Step-by-step Derivation:

1. Identify the given values:
Mass of the first particle:

$m_1=1\text{g}$

Velocity of the first particle:

${\overrightarrow{v}}_1=3î-2ĵ$

Mass of the second particle:

$m_2=2\text{g}$

Velocity of the second particle:

${\overrightarrow{v}}_2=4ĵ-6\mathrm{k̂}$

2. Apply the Principle of Conservation of Linear Momentum:
In a perfectly inelastic (glued) collision, the two particles stick together after the impact to form a combined particle. Since no external forces act on the system, the total initial momentum is equal to the total final momentum:

${\overrightarrow{P}}_{\text{initial}}={\overrightarrow{P}}_{\text{final}}$

$m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2=(m_1+m_2){\overrightarrow{v}}_f$

where ${\overrightarrow{v}}_f$ is the velocity of the newly formed particle.

3. Calculate the total initial momentum:

$m_1{\overrightarrow{v}}_1=1·(3î-2ĵ)=3î-2ĵ$

$m_2{\overrightarrow{v}}_2=2·(4ĵ-6\mathrm{k̂})=8ĵ-12\mathrm{k̂}$

Adding these momentum vectors together:

${\overrightarrow{P}}_{\text{initial}}=(3î-2ĵ)+(8ĵ-12\mathrm{k̂})$

${\overrightarrow{P}}_{\text{initial}}=3î+6ĵ-12\mathrm{k̂}$

4. Determine the velocity vector of the combined particle:
The combined mass of the system is:

$M=m_1+m_2=1+2=3\text{g}$

Using the momentum conservation equation:

$3{\overrightarrow{v}}_f=3î+6ĵ-12\mathrm{k̂}$

Dividing both sides by 3:

${\overrightarrow{v}}_f=1î+2ĵ-4\mathrm{k̂}$

5. Find the magnitude of the final velocity:

$\left|{\overrightarrow{v}}_f\right|=\sqrt{1^2+2^2+{(-4)}^2}$

$\left|{\overrightarrow{v}}_f\right|=\sqrt{1+4+16}$

$\left|{\overrightarrow{v}}_f\right|=\sqrt{21}\approx 4.58{\text{ms}}^{-1}$

Rounding to one decimal place, the speed of the combined particle is:

$4.6{\text{ms}}^{-1}$