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blondinia [14]
3 years ago
9

A of mass 5.0 kilograms moving at 20. meters per second collides with ball B of unknown mass moving at 10. meters per second in

the same direction. After the collision, ball A moves at 10. meters per second and ball B at 15 meters per second, both still in the same direction. What is the mass of ball B
Physics
1 answer:
barxatty [35]3 years ago
7 0

Answer:

Mass of B is 10 kg

Explanation:

M1v1a + m2v1b = m1v2a + m2v2b

Let m2 = x

5*20 + x10 = 5*10 + x15

100 + x10 = 50 + x15

50 = x5

X= 10

Mass of B is 10 kg

M1 = mass of A

V1a = initial velocity of A

V2a = final velocity of A

M2 = mass of B

V1b = initial velocity of B

V2b = final velocity of B

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Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

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Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

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