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jasenka [17]
3 years ago
10

A 5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at re

st.
(a) Find the velocity of each object after the collision. cm/s (5.00 g object) cm/s (10.0 g object)
(b) Find the fraction of the initial kinetic energy transferred to the 10.0 g object. %
Physics
1 answer:
marusya05 [52]3 years ago
3 0

Answer: a) 6.67cm/s b) 1/2

Explanation:

According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".

Let m1 and m2 be the masses of the bodies

u1 and u2 be their velocities respectively

m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s

Since momentum = mass × velocity

The conservation of momentum of the body will be

m1u1 + m2u2 = (m1+m2)v

Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.

5(20) + 10(0) = (5+10)v

100 + 0 = 15v

v = 100/15

v = 6.67cm/s

Therefore the velocity of each object after the collision is 6.67cm/s

b) kinectic energy of the 10.0g object will be 1/2MV²

= 1/2×10×6.67²

= 222.44Joules

kinectic energy of the 5.0g object will be 1/2MV²

= 1/2×5×6.67²

= 222.44Joules

= 111.22Joules

Fraction of the initial kinetic transferred to the 10g object will be

111.22/222.44

= 1/2

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Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill
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Answer:

a)   m =1  θ = sin⁻¹  λ  / d,  m = 2        θ = sin⁻¹ ( λ  / 2d) ,   c)     m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

         d sin θ = m λ

the maximum for m = 1 is at the angle

          θ = sin⁻¹  λ  / d

the second maximum m = 2

          θ = sin⁻¹ ( λ  / 2d)

the third maximum m = 3

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the fourth maximum m = 4

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b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

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       x = pi a sin θ /λ

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with the values ​​of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

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we divide the two expressions

                       d / a = m

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                    m = 3

order three is no longer visible

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