A glider of mass 0.160 kg is moving to the right on a frictionless, horizontal air track with a speed of 0.900 m/s . It has a head-on collision with a glider 0.310 kg that is moving to the left with a speed of 2.18 m/s . Suppose the collision is elastic. Find the final velocity (magnitude and direction) of each glider.

- V_a2 is 3.156 m/s in the left direction

- V_b2 is 0.076 m/s in the left direction

Explanation:

Since the 2 gliders are moving on a frictionless surface, then there is no net external force on the system and thus the collision is elastic;

In elastic collisions, the relative velocities before and after collision have the same values.

Thus,

V_a1 - V_b1 = V_b2 - V_a2

Where,

V_a1 is initial velocity of first glider

V_a2 is final velocity of first glider

V_b1 is initial velocity of second glider

V_b2 is final velocity of second glider

From the question,

V_a1 = 0.9 m/s

V_b1 = 2.18 m/s

Thus,

0.9 - (-2.18) = V_b2 - V_a2

V_b2 - V_a2 = 3.08 m/s - - - - (1)

Now, the collision equation is written as;

m_a•V_a1 + m_b•V_b1 = m_a•V_a2 + m_b•V_b2

We are given that;

m_a = 0.16kg

m_b = 0.31 kg

Thus, plugging in the values,

(0.16 x 0.9) + (-0.31 x 2.18) = 0.16•V_a2 + 0.31•V_b2

0.144 - 0.6758 = 0.16•V_a2 + 0.31•V_b2

0.16•V_a2 + 0.31•V_b2 = -0.5318

Divide through by 0.16;

V_a2 + 1.9375•V_b2 = -3.324 - - - (2)

Let's add eq1 and 2 to get;

V_b2 + 1.9375•V_b2 = 3.08 - 3.324

2.9375V_b2 = -0.224

V_b2 = -0.224/2.9375 = - 0.076 m/s

Negative sign means it is in an opposite direction which in this case is to the left.

Thus, V_b2 is 0.076 m/s in the left direction

Now, let's find V_b2.

Let's put -0.076 m/s for V_b2 in eq 1;

Thus,

-0.076 m/s - V_a2 = 3.08 m/s

V_a2 = -0.076 m/s - 3.08 m/s

V_a2 = -3.156 m/s

Negative sign means it is in an opposite direction which in this case is to the left.

Thus, V_a2 is 3.156 m/s in the left direction

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.