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Alekssandra [29.7K]
2 years ago
7

Why is the contribution of the wavelets lying on the back of secondary wave front zero?​

Physics
2 answers:
Tresset [83]2 years ago
3 0

Answer:

The contribution of the wavelets lying on the back of the wave front is zero because of something known as the Obliquity Factor. It is assumed that the amplitude of the secondary wavelets is not independent of the direction of propagation, Sources: byju's.com

Anna [14]2 years ago
3 0

According to the Voigt and Kirchhoff's law for wavelets, the contribution of a wavelet in any direction making an angle x with the normal to the wavelet is proportional to {1+cos(x)/2)}

For wavelet lying on the back of secondary wavefront,

  • x = 180°

→ So, the proportional factor:

→ {1+cos(180°)/2}

→ (1-1/2)

→ (0/2)

→ 0

That's why the contribution of the wavelets lying on the back of secondary wave front is zero

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Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0
3 years ago
Read 2 more answers
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