# Calculate the Velocity of the Centre of Mass of the System of Particles Shown in Figure. - Physics

Sum

Calculate the velocity of the centre of mass of the system of particles shown in figure.

#### Solution

From the figure, the velocities of different masses can be written as:

$\text{For m}_1 = 1 . 0 \text{kg},$

$\text{ Velocity, } \vec{v}_1 = \left( - 1 . 5 \cos 37^\circ \hat i - 1.5 \sin 37^\circ\hat j\right) = - 1 . 2 \hat i- 0.9 \hat j$
$\text{For m}_2 = 1 . 2 \text{kg},$

$\text{Velocity}, \vec{v}_2 = 0 . 4 \vec{j}$

$\text{For m}_3 = 1 . 5 \text{kg,}$

$\text{Velocity,} \vec{v}_3 = - 1 . 0 \cos 37^\circ \hat i0 + 1.0 \sin 37^\circ\hat j$

$\text{ For m}_4 = 0 . 50 \text{kg},$

$\text{ Velocity,} \vec{v}_4 = 3 . 0 \hat i$

$\text{ For m}_5 = 1 . 0 \text{ kg},$

$\text{ Velocity }, \vec{v}_5 = 2 . 0 \cos 37^\circ \hat i - 2 . 0 \sin 37^\circ \hat j$

$(\cos 37^\circ = \frac{4}{5} \text{ and } \sin 37^\circ \ = \frac{3}{5})$

$V_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3 + m_4 \vec{v}_4 + m_5 \vec{v}_5}{m_1 + m_2 + m_3 + m_4 + m_5}$

$= \frac{1}{1 . 0 + 1 . 2 + 1 . 5 + 1 . 0 + 0 . 50}^\left[ 1 . 0\left( - 1 . 5 \times \frac{4}{5} \vec{i} - 1 . 5 \times \frac{3}{5} \vec{j} \right) + . . . - 2 . 0 \times \frac{3}{5} \vec{j} \right]$

On solving the above equation, we get:
Vcm is 0.20 m/s , at 45° below the direction, towards right.

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 9 Centre of Mass, Linear Momentum, Collision
Q 7 | Page 160