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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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A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling
sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

P_{1} = P_{2}

m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

v_{1_{i}}: is the initial velocity of the ball 1 = 6.20 m/s

v_{2_{i}}: is the initial velocity of the ball 2 = 0 (it is at rest)

v_{1_{f}}: is the final velocity of the ball 1 =?

v_{2_{f}}: is the initial velocity of the ball 2 =?

m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}

v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}} (1)        

Now, by conservation of kinetic energy (since they collide elastically):

\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}          

m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}  (2)

By entering equation (1) into (2) we have:

m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}    

0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}            

By solving the above equation for v_{2_{f}}:

v_{2_{f}} = 3.1 m/s

Now, v_{1_{f}} can be calculated with equation (1):

v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s

The minus sign of v_{1_{f}} means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!                  

5 0
3 years ago
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