Determine the velocity of the centre of mass for the system of...

Let the right direction be the positive x-axis and the upward direction be the positive y-axis. The total mass of the system is:

$M = 1.0 + 1.2 + 1.5 + 0.50 + 1.0 = 5.2 \text{ kg}$

The total momentum in the x-direction is:

P_x = -1.0(1.5 \cos 37^{}^{{}^{\circ}}) + 0 - 1.5(1.0 \cos 37^{}^{{}^{\circ}}) + 0.50(3.0) + 1.0(2.0 \cos 37^{}^{{}^{\circ}})

Using \cos 37^{}^{{}^{\circ}} = 0.8, we get:

$P_x = -1.2 + 0 - 1.2 + 1.5 + 1.6 = 0.7 \text{ kg m/s}$

The total momentum in the y-direction is:

P_y = -1.0(1.5 \sin 37^{}^{{}^{\circ}}) + 1.2(0.4) + 1.5(1.0 \sin 37^{}^{{}^{\circ}}) + 0 - 1.0(2.0 \sin 37^{}^{{}^{\circ}})

Using \sin 37^{}^{{}^{\circ}} = 0.6, we get:

$P_y = -0.9 + 0.48 + 0.9 + 0 - 1.2 = -0.72 \text{ kg m/s}$

The components of the velocity of the centre of mass are:

$v_{cm,x} = \dfrac{P_x}{M} = \dfrac{0.7}{5.2} \approx 0.135 \text{ m/s}$

$v_{cm,y} = \dfrac{P_y}{M} = \dfrac{-0.72}{5.2} \approx -0.138 \text{ m/s}$

The magnitude of the velocity is:

$v_{cm} = \sqrt{v_{cm,x}^2 + v_{cm,y}^2} = \sqrt{(0.135)^2 + (-0.138)^2} \approx 0.193 \text{ m/s}$

Rounding to two significant figures, $v_{cm} \approx 0.20 \text{ m/s}$.

The angle $\theta$ with the positive x-axis is given by:

$\tan \theta = \dfrac{v_{cm,y}}{v_{cm,x}} = \dfrac{-0.72}{0.70} \approx -1.03$

This corresponds to an angle of approximately 45^{}^{{}^{\circ}} below the positive x-axis (towards the right).