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dmitriy555 [2]
3 years ago
14

The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick

together and continue on with velocity vf. Which of these statements is true?
a. vf= v2
b. vf is less than v2.
c. vf is greator than v2, but less than v1.
d. vf= v1
Physics
1 answer:
Travka [436]3 years ago
4 0

Answer:

c. vf is greator than v2, but less than v1

Explanation:

The principle of conservation of linear momentum states that when two or more bodies act upon one another, their total momentum remains constant.

In a system of colliding bodies the total momentum of the system just before the collision is the same as the total momentum just after the collision.  

Collisions in which the kinetic energy is conserved are called elastic collision.

Collisions in which the kinetic energy is not conserved are called inelastic collisions.  If the two objects stick together after the collision and move with a common velocity, the collision is said to be perfectly inelastic.

<em>The above scenario is a perfectly inelastic collision. The initial velocity of particle 1 was greater than particle 2 before collision. After collision, its velocity will reduce to a final velocity vf as it transfers some of its kinetic energy to particle 2; whereas, the velocity of particle 2 will increase to a final velocity vf as it absorbs some of the kinetic energy of particle 1.</em>

Therefore,

a. vf = v2 is wrong because vf is greater than v2

b. vf is less than v2 is wrong because vf is greater than v2

c. vf is greater than v2, but less than v1 is correct.

d. vf = v1 is wrong because vf is less than v1

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The current density is equal:

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J=4A/(π*1*10^-6m^2)=1.27 *10^6 A/m^2

The resistence of the wire is given by:

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Then we have:

σ= L/(R*A)=1/(5.5*10^-3*π*1*10^-6)=57.9 *10^6 Ω*m

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