If two billiard balls collide, what happens to the systems overall momentum ?

1 answer:

4 0

Remains same. Here according to law of conservation of momentum, the system have momentum conserved and remains constant.
Given by the popular equation,
m1v1=m2v3
m1=Mass of billiard ball 1
m2= Mass of the 2nd ball
v1,v2 are the velocity of the ball 1,2 respectively.

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Further Explanation:

Consider the pressure of the gas to be constant.

The change in the volume of the gas when the temperature of the gas is varied by keeping the pressure of the gas at a constant value is defined by the Charles' Law.

Concept:

According to the Charles law, the volume of the gas is directly proportional to the temperature of the gas at constant pressure.

The Charles' law can be stated as:

$\fbox{\begin\\V\propto T\\\end{minispace}}$

The above expression can we written as.

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

Convert the temperature of the gas into kelvin.

$T=273+T^\circ\text{C}}}$

Here, T is the temperature in kelvin and $T^\circ\text{C}}}$ is the temperature in centigrade.

The initial temperature of the gas is $50^\circ\text{C}$ . The temperature of the gas in kelvin is.

$\begin{aligned}{T_1}&=273+50\\&=323\,{\text{K}}\\\end{aligned}$

The final temperature of the gas is $500^\circ\text{C}$ . The temperature in kelvin is.

$\begin{aligned}{T_2}&=273+500\\&=773\,{\text{K}}\\\end{aligned}$

Substitute the values of temperature and volume in the expression of the Charles' Law.

$\begin{aligned}{V_2}&=\frac{{{T_2}}}{{{T_1}}}{V_1}\\&=\frac{{773\,{\text{K}}}}{{323\,{\text{K}}}}\left({2.33\,{\text{L}}}\right)\\&=5.576\,{\text{L}}\\\end{aligned}$