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KiRa [710]
3 years ago
8

If two objects collid and one is initially at rest, is it possible for both to be at rest after the collision?

Physics
1 answer:
Ludmilka [50]3 years ago
5 0

Answer:

<em>Is it NOT possible for both particles to be at rest after the collision</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum</u>

The total linear momentum of a system of particles is conserved regardless of their internal interactions while no external force is applied to the system. If m_1 and m_2 are the masses of two particles, and only one of them is at rest, there is a non-zero total linear momentum, i.e.

P_1=m_1v_1+m_2v_2

if at least one of the speeds is not zero, the total initial linear momentum is not zero.

If after the collision, both particles remain at rest, both speeds will be zero and the total linear momentum won't be conserved.

So, is it NOT possible for both particles to be at rest after the collision

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Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength
soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ  + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

V = \sqrt{\frac{2K.E}{m} }\\\\V =  \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5}  \ m/s

the shortest de Broglie wavelength for the electrons is given by;

\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0
4 years ago
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