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AleksandrR [38]
3 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. Perfectly Inelastic Collisions Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial velocity of the ball A is 6 meters per second to the right, and the initial velocity of the ball B is 12 meters per second to the left.
Physics
2 answers:
ratelena [41]3 years ago
8 0
1. The balls move to the opposite direction but the same speed. This represents Newton's third law of motion.
2. The total momentum before and after the collision stays constant or is conserved.
3. If the masses were the same, the velocities of both balls after the collision would exchange.
4 and 5. Use momentum balance to solve for the final velocities.
Citrus2011 [14]3 years ago
5 0

Answer:

Explanation:

1) Initial momentum of A = 18 kg m/s towards the right ; Initial momentum of B is 18 kg m/s towards the left. Total momentum of the system = 0.

After the collision momentum of A = 18 kg m/s towards the left ; momentum of

B = 18 kg m/s towards the right. Total momentum = 0

. Change in momentum of A = 18 - (-18) = 36 kg m/s

Change in momentum of B - 18 - 18 = -36.

Total change in momentum= 0

Change in momentum in both occur due to two forces created at the point of interaction . They are called action and reaction forces according to newton's third law.

2. Total momentum remains zero before and after the collision.

3. Then, they would have bounced back with equal velocity to conserve momentum

4. Applying law of conservation of momentum

m₁u₁ +m₂u₂ = m₁v₁ +m₂v₂

  10x3 +5x-3 = 10x2 + 5xv₂

     v₂ = -1 m/s

Ball B will move towards the left.

5. If the collision is perfectly elastic, there will be exchange of velocity ie B will have twice the velocity than that of A after the collision.

For perfectly inelastic collision

V = (m₁v₁ + m₂v₂)/( m₁ + m₂) = 7 x6 + 2 x -12/ 7+2 = 2 m/s towards the right.

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A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

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initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

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(c)

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the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

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kipiarov [429]

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