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Yuki888 [10]
3 years ago
11

Let two objects of equal mass m collide. Object 1 has initial velocity v, directed to the right, and object 2 is initially stati

onary. Now suppose that the collision is perfectly inelastic. What are the velocities v1 and v2 of the two objects after the collision?
Physics
1 answer:
irakobra [83]3 years ago
5 0

Answer:

v₁ = v₂ = v/2

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

        m_{1}* v = m_{1} *v_{1f}  + m_{2}*v_{2f} (1)

  • If the collision is perfectly inelastic, both masses stick together after the collision, which means that both masses will have the same final speed:

        v_{1f} = v_{2f} (2)

  • We know that both masses are equal each other:

        m_{1} =m_{2} = m  (3)

  • From (1), (2) and (3), we can solve for v₁f = v₂f, as follows:

        m v = 2*m *v_{1f}

        ⇒ v_{1f} = v_{2f} =\frac{v}{2}

  • The velocities of both objects after the collision are the same, and equal to v/2.
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A 1020 kg boat is traveling at 110 km/h when its engine is shut off. The magnitude of the frictional force k between boat and wa
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An exact solution will require calculus, since the acceleration is not constant.


M*dV/dt = -fk = -75V
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Since you have separated variables to opposite sides, the differential equation is easily integrated. 



Therefore,

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Then we have that,

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4 0
3 years ago
When light with a wavelength of 238 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic ener
erastova [34]

Answer:

The wavelength is 173 nm.

Explanation:

This kind of phenomenon is known as photoelectric effect, it occurs when photons of light inside the metal surface and if they have the right amount of energy electrons absorb it and got expelled from the metal as photo electrons. The maximum kinetic energy of that photo electrons is given by the expression:

K_{max} =E_{photon} - \Phi (1)

With E the energy of the photon and Φ the work function of the material. The work function is a value characteristic of each material and is related with how much the electron is attached to the material, the energy of the photon is the Planck's constant (h=6.63\times10^{-34}) times the frequency of light (\nu) , then (1) is:

K_{max} =h\nu - \Phi (2)

The frequency of an electromagnetic wave is related with the wavelength (\lambda) by:

\nu=\frac{c}{\lambda} (3)

with c the velocity of light (c=3.0\times10^{8})

Using (3) on (2):

K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \Phi:

\Phi=\frac{hc}{\lambda}-K_max=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{238\times10^{-9}}-3.13\times10^{-19}

\Phi=5.23\times10^{-19} J

That's the work function of the metal we're dealing. So now if we want to know the wavelength to obtain the double of the kinetic energy we use:

2K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \lambda:

\lambda = \frac{hc}{2K_{max}+\Phi}=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{2(3.13\times10^{-19})+5.23\times10^{-19}}=1.73\times10^{-7}

\lambda=173 nm

3 0
4 years ago
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