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2 years ago

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A group of particles of total mass 35 kg has a total kinetic energy of 313 J. The kinetic energy relative to the center of mass

is 81 J. What is the speed of the center of mass?

1 answer:

lutik1710 [3]2 years ago

5 0

Answer:

The speed of center of mass is 3.64 m/s.

Explanation:

Given that,

Total mass of the group of particles, m = 35 kg

Total kinetic energy of group of particles, $K_T= 313 J$

The kinetic energy relative to the center of mass is, K = 81 J

We need to find the speed of center of mass. We know that the total kinetic energy is equal to the sum of rotational kinetic energy and the rotational kinetic energy.

$K_T=E+K\\\\E=K_T-K$

Since, $E=\dfrac{1}{2}mv_c^2$ , $v_c$ is the speed of center of mass

$\dfrac{1}{2}mv_c^2=K_T-K\\\\v_c=\sqrt{\dfrac{2(K_T-K)}{m}} \\\\v_c=\sqrt{\dfrac{2(313-81)}{35}}\\\\v_c=3.64\ m/s$

So, the speed of center of mass is 3.64 m/s.

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Answer:

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Solution:

As per the question:

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Average atomic weight is given as:

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$A_{Ag}$ = 107

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$A_{Pd}$ = 106

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In the similar way, average density is given as:

$\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}$

$\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}$

Therefore, edge length is given by eqn (1) as:

$a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm$

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