The velocities of three particles of masses 20g, 30g and 50g are 10î ,10 , ĵ , and 10k̂ respectively. The velocity of the centre of mass of the three particles is

To find the velocity of the centre of mass of a system of particles, we use the standard formula for the velocity of the centre of mass ($v_{\mathrm{cm}}$):

$v_{\mathrm{cm}}=\frac{m_1{v}_1+m_2{v}_2+m_3{v}_3}{m_1+m_2+m_3}$

Let's identify the given values for each of the three particles from the problem statement:
For the first particle:
Mass ($m_1$) = 20 g
Velocity ($v_1$) = 10î

For the second particle:
Mass ($m_2$) = 30 g
Velocity ($v_2$) = 10ĵ

For the third particle:
Mass ($m_3$) = 50 g
Velocity ($v_3$) = 10k̂

Now, let's calculate the total mass of the system ($M$):

$M=m_1+m_2+m_3=20+30+50=100\text{g}$

Next, we calculate the numerator, which represents the total momentum (expressed in terms of mass in grams and velocity):

$m_1{v}_1+m_2{v}_2+m_3{v}_3=20\left(10î\right)+30\left(10ĵ\right)+50\left(10\mathrm{k̂}\right)$

$=200î+300ĵ+500\mathrm{k̂}$

Finally, we divide this by the total mass to obtain the velocity of the centre of mass:

$v_{\mathrm{cm}}=\frac{200î+300ĵ+500\mathrm{k̂}}{100}$

$v_{\mathrm{cm}}=2î+3ĵ+5\mathrm{k̂}$

Therefore, the velocity of the centre of mass of the three particles is 2î + 3ĵ + 5k̂.