Question

Three particles of equal masses travelling with velocities of 10m/s, 20 m/s
and 30 m/s , respectively, along x-axis , at an angle of 30° to the direction
of positive x-axis and y-axis (as shown in Fig.) collide simultaneously and
get sticked to each other. combined particle will move at angle a with x-axis where​

Answer 1

Answer:

The Particle will move along x-axis at

$\frac{10}{3}\sqrt{20+2 \sqrt{3}m/s$

Explanation:

Particle (P1) long x - axis.

Velocity of particle ( P1 ) = 10 m/s

Velocity of particle ( P2 ) = 20 m/s

Velocity of particle ( P3 ) = 30 m/s

Formula:

           Linear Momentum (p)= Mass(m) ×Velocity(v)

Momentum of particle ( P1 ) along positive x-axis = m(10i) = (10m) i

Momentum of particle( P2 ) at an angle 30° to the direction of positive x-axis = m(20cos30°i + 20sin30°j) =(10√3m)i + (10m)j

Momentum of particle( P3 ) along y-axis= m(30j) = (30m)j

After collision, Particle sticked together. let their velocity of system of particles after collision is v. hence final momentum wuld be:

            Final Linear Momentum (Pf )= (m + m + m)v = 3mv

According to Law of Conservation of Linear Momentum,

                    initial momentum = final momentum

$P_{1} +P_{2} +P_{3} =P_{f}\\Putting \ Values \ and \ simplyfiying,\\(10m)i + (10\sqrt{3}m)i + (10m)j + (30m)j = 3mv\\(10 + 10\sqrt{3})i + 40j = 3v\\v = (\frac{10+10\sqrt{3}}{3})i+(\frac{40}{3})j\\v=\frac{10}{3}[(1+\sqrt{3})i+4j]$

Formula:                

                  Magnitude of Vector |v| = √(a^2+b^2)

$Magnitude \ of \ Velocity \ |v| = \frac{10}{3}\sqrt{(1+\sqrt{3})^2+4^2}\\ =\frac{10}{3}\sqrt{[(1^2+3^2+2\sqrt{3})+16]$

$=\frac{10}{3}\sqrt{1+3+16+2\sqrt{3}}\\Magnitude \ of \ Velocity \ |v|=[\frac{10}{3}\sqrt{20+2 \sqrt{3}m/s$

HOPE YOU FIND IT HELPFUL :)