Question

Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicular to each other (take A initially moving to the right and B initially moving up). Use momentum conservation (make a complete Momentum Chart) to find the velocity (magnitude and direction with-respect-to the velocity asteroid A had before the collision) of the asteroids after the collision.

Answer 1

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $10^5$

So, $V_x = \frac{10^5}{3000}$  = $\frac{100}{3}$  m/s

and $V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $\sqrt{V_x^2 + V_y^2}$

                                   = $\sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2}$

                                   = 62.89 m/s

And direction is

tan θ = $\frac{V_y}{V_x}$     = 1.6

therefore, $\theta = \tan^{-1}1.6$

                   = $58 ^{\circ}$  from x-axis