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solniwko [45]
2 years ago
13

Object A of mass M is moving east at speed v. It collides with object B of mass 2M that was initially at rest. The motion of the

objects before and after the collision is along the same line. After the collision, object A is moving west at a speed of v/3. What is the speed of object B immediately after the collision?
Physics
1 answer:
Firlakuza [10]2 years ago
3 0

Answer:

v_B=\frac{v}{3}

Explanation:

Given that:

mass of object A, m_A=M

mass of object B, m_B=2M

speed of object A, v_A=v

So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.

m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B

M\times v+0 = M\times \frac{v}{3}+2M\times v_B

2M\times v_B= \frac{2M\times v}{3}

v_B=\frac{v}{3}

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he first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold
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Answer:

Relative population is  2.94 x 10⁻¹⁰.

Explanation:

Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.

The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:

\frac{N_{1} }{N_{2} } =  \frac{g_{1}e^{\frac{-E_{1} }{KT} }  }{g_{2}e^{\frac{-E_{2} }{KT} }}

\frac{N_{1} }{N_{2} } = \frac{g_{1}e^{\frac{-(E_{1}-E_{2})  }{KT} }  }{g_{2}}

Here g₁ and g₂ be the degeneracy at two levels, K is Boltzmann constant and T is equilibrium temperature.

Put 1 for g₁, 3 for g₂, -19.82 ev for (E₁ - E₂) and 8.6x10⁵ ev/K for K and 10000 k for T in the above equation.

\frac{N_{1} }{N_{2} } = \frac{1\times e^{\frac{-(-19.82)}{8.6\times 10^{-5}\times 10000} }  }{3}

\frac{N_{1} }{N_{2} } = 3.4 x 10⁹

\frac{N_{2} }{N_{1} } =  2.94 x 10⁻¹⁰

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The temperature will remain constant, at around 100 C, and the volume of water in the pot will decrease, as it turns into steam and floats away from the pot.
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Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms o
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U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

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Given:

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- The total mechanical energy E_top at the top position h_max in terms of the available variables.

Solution:

- First we need to determine the types of Energy that are in play:

- The Elastic potential Energy E_p in a spring is given:

                              E_p: 0.5 * k * (ext)

- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:

             ext = Equilibrium + distance pulled - h_max = (x + d - h_max)

Hence, the elastic potential energy will be:

                              E_p = 0.5 * k *(x + d - h_max)^2

- The gravitational potential energy E_g is given by:

                              E_g = m*g*h_max

Where, bottom most position is taken as reference (datum).

- The kinetic Energy E_k is given by:

                              E_k = 0.5*m*v_top^2

- Since we know that the maximum height is reached when velocity is zero

Hence,                   E_k = 0.5*m*0^2 = 0.

The total Energy of the system U is sum of all energies and play:

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                               U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

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Answer:

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