Please help I don't know what the answer is
1 answer:
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1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we call
the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
(1)
where
is the mass of ball A
is the mass of ball B
is the initial velocity of ball A
is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)
If we solve (1) to find, we find that the final velocity of the balls is
and the positive sign means the two balls are going to the right.
2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
where
is the final velocity of ball A
is the final velocity of ball B
If we solve simultaneously the two equations, we find:
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.
3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
and the total momentum after the collision is:
If we call
where
If we solve (1) to find
and the positive sign means the two balls are going to the right.
2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
where
If we solve simultaneously the two equations, we find:
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.
3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
and the total momentum after the collision is:
This question is incomplete, the complete question;
you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.
Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity
Options;
a) 200 nm; 0.9 mW
b) 100 nm, 0.0059 mW
c) 200 nm; 0 mW
d) 100 nm; 0.9 mW
e) 200 nm; 0.0059 mW
Answer:
the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;
100 nm; 0.0059 mW
Option b) 100 nm, 0.0059 mW is the correct answer
Explanation:
Given that the instrument here is an interferometer.
Maximum intensity is obtained when the two waves are exactly in phase.
that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.
The phase factor of this point is taken as ∅ = 0
Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π
This corresponds to a path difference between the two waves as half of the wavelength. λ/2
The light gets reflected from the mirror.
Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)
hence, to get a path difference of λ/2 the mirror should move half of this distance only
so, the mirror should move;
= λ/4
here, wavelength is 400nm
the length moved by the mirror = 400/4 = 100 nm
The intensity is given by the equation;
=
1 +
2 + 2√
1
2cos(∅)
where
1 = 2.25 mW
2 = 2.025 mW
∅ = π
so we substitute
= 2.25 + 2.025 - 2√(2.25 × 2.025)
= 4.275 - 4.26907
= 0.0059
Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;
100 nm; 0.0059 mW
Option b) 100 nm, 0.0059 mW is the correct answer