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arlik [135]
2 years ago
5

Imagine two billiard balls on a pool table. Ball A has a mass of 2 kilograms and ball B has a mass of 3 kilograms. The initial v

elocity of ball A is 9 meters per second to the right, and the initial velocity of the ball B is 6 meters per second to the left. The final velocity of ball A is 9 meters per second to the left, while the final velocity of ball B is 6 meters per second to the right.
1. Explain what happens to each ball after the collision. Why do you think this occurs? Which of Newton’s laws does this represent?

2. What can you say about the total momentum before and after the collision?

3. What do you think would happen to the velocity of each ball after the collision if the masses and initial velocities of each ball were the same?

4. The mass of ball A is 10 kilograms and the mass of ball B is 5 kilograms. If the initial velocity is set to 3 meters per second for each ball, what is the final velocity of ball B if the final velocity of ball A is 2 meters per second? Use the elastic collision equation to find the final velocity of ball B. Assume ball A initially moves from right to left and ball B moves in the opposite direction. Identify each mass, velocity, and unknown. Show your work, including units, and indicate the direction of ball B in your answer.

5. If the mass of each ball were the same, but the velocity of ball A were twice as much as ball B, what do you think would happen to the final velocity of each ball after the collision? To answer this question, create a hypothesis in the form of an if-then statement. The “if” is the independent variable, or the thing that is being changed. The “then” is the dependent variable, or what you will measure as the outcome.
(Please show your work if you can)
Physics
1 answer:
Orlov [11]2 years ago
3 0
1. After the collision, their velocities have switched. This is Newton's third law of motion.

2. The total momentum is conserved.

3. The same thing would happen if the collision is totally elastic.

For 4 and 5.
Using the conservation of momentum equation
m1v1 + m2v2 = m1'v1' + m2'v2'<span />
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Answer:

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External pressure is given as  1.00atm , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence,  V_i = 0.

To find the volume of the products, we need to first find the amount of moles of the product made from  2.40_gKNO_3, using the molar mass of  KNO_3  which is 101.1032 g/mol  

2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3

Now let us convert moles of  KNO_3  into moles of CO_2 and N_2  using the stoichiometric ratios from our balanced equation of the reaction.

0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2

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0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2

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1.567_gCO_2 . {\frac{1L}{1.830_g}} = 0.856LCO_2

0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2

Summing up the two volumes, we get the final volume

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Plugging everything into the w_{pv} equation, we get:

w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm

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