Answer to Question #96540 in Classical Mechanics for Tella David

Question

How do you say a vehicle of mass 3000 0 0 moving with a velocity of 15 m per second at the degree of 90° right with another vehicle of mass 400000 m moving with a velocity of 12 metre per second at a degree of 120 degree calculate the magnitude and direction of the common velocity with which the velocity begins to move immediately after

Answer 1

From the conservation of momentum:

$$m_1\bold{v_1}+m_2\bold{v_2}=(m_1+m_2)\bold{v}$$

$$m_1\bold{v_1}=3000(0,15)=(0,45000)\frac{kgm}{s}$$

$$m_2\bold{v_2}=4000(12\cos{120},12\sin{120})=(-24000,41569)\frac{kgm}{s}$$

$$v_x=\frac{-24000}{3000+4000}=-3.43\frac{m}{s}$$

$$v_y=\frac{45000+41569}{3000+4000}=12.37\frac{m}{s}$$

The magnitude:

$$v=\sqrt{(-3.43)^2+12.37^2}=12.8\frac{m}{s}$$

The direction:

$$\theta=180\degree-\arctan{\frac{12.37}{3.43}}=105.5\degree$$