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Kryger [21]
2 years ago
11

Two objects of the same mass travel in opposite directions along a horizontal surface. Object X has a speed of 5ms and object Y

has a speed of 5ms, as shown in the figure. After a period of time, object X collides with object Y. In scenario 1, the objects stick together after the collision. In scenario 2, the objects do not stick together after the collision.
Which of the following claims is true regarding how the outcome of scenario 1 is different from the outcome of scenario 2?
Physics
2 answers:
Papessa [141]2 years ago
8 0

Answer:

1

Explanation:

Because theyre heading opposite directions

alexdok [17]2 years ago
8 0
Since both objects are travelling in opposite directions at the same speed but eventually colliding with each other hence it is obvious that they are travelling in a circle.

Now, this question is regarding the conversation of momentum - elastic / in elastic collision.

Assuming the balls are in a closed system(an assumption that is consistent in your syllabus,unless stated otherwise)as the ball collides, momentum is conserved but some of the energy might be lost due to the collision either through heat / sound.

Summary:

3 types of collision.

1st type: Elastic collision
- No loss in Kinetic Energy
- No loss in momentum
- Balls do not stick together
Example: Bouncing Basketball, the ball comes back to your hand at the same height.

2nd type: Inelastic collision
- Loss in kinetic energy, (heat energy when in contact)
- No loss in momentum
- Balls do not stick together
Example: Bomb explosion.

3rd type: Perfectly Inelastic collision
- Maximum amount of KE is loss
- No loss in momentum
- Balls stick together
Example: Ballistic Pendulum
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GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

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Let v is the resultant of velocity. It is given by :

v=\sqrt{v_1^2+v_2^2}

v=\sqrt{5^2+8^2}=9.43\ m/s

Let E is the kinetic energy of the plane. It is given by :

E=\dfrac{1}{2}mv^2

E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2

E = 133.38 Joules

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2 years ago
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PART A)

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here we have

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now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

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Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

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W = e(0 - 0) = 0

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PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

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KE = 4.55 \times 10^{-27} kg

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PART C)

It will reach the origin

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Answer:

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