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irina1246 [14]
3 years ago
15

1. If an object of mass m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, th

en the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor of ___. The new velocity will be ___v.
2. If an object of mass m collides and velocity v collides inelastically with an object of mass 4m that is initially at rest, then the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor of ___. The new velocity will be ___v.

3. If an object of mass 3m collides and velocity v collides inelastically with an object of mass 4m that is initially at rest, then the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor if ___. The new velocity will be ___v.

4. If an object of mass 5m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, then the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor if ___. The new velocity will be ___v.
Physics
1 answer:
Archy [21]3 years ago
3 0

Answer:

1a). 4 times. 1b) 4. 1c) 1/4.

2a) 5 times. 1b) 5 1c) 1/5

3a) 7/3 times 3b) 7/3 3c) 3/7

4a) 8/5 times 4b) 8/5 4c) 5/8

Explanation:

1) Assuming no external forces acting during the collision, total momentum must be conserved, so the following general expression applies:

m₁*v₁₀ + m₂*v₂₀ = m₁*v₁f  + m₂*v₂f (1)

If we assume that the collision is perfectly inelastic, this means that both masses stick together after the collision, so v₁f = v₂f.

If m₂ is initially at rest, ⇒ v₂₀ = 0.

Replacing in (1) we get the expression of vf as a function of v₁₀, as follows:

vf = v₁₀*(m₁/(m₁+m₂)

So, for the four cases we have the following:

1) initial mass = m

  final mass = m+3m = 4 m

⇒final mass / initial mass = 4

vf = v₀* (m/4m) = v₀/4  ⇒v₀/vf = 4

So, the velocity of the system will decrease by a factor of 4. The new velocity will be vf= v₀/4.

2) Applying the same considerations, we get:

2a)  final mass / initial mass = 5

2b) vf = v₀* (m/5m) = v₀/5  ⇒v₀/vf = 5

2c) vf = v₀/5

3) Applying the same considerations, we get:

3a)  final mass / initial mass = 7/3

3b) vf = v₀* (3m/7m) =3/7* v₀  ⇒v₀/vf = 7/3

3c) vf = 3/7*v₀

4) Applying the same considerations, we get:

4a)  final mass / initial mass = 8/5

4b) vf = v₀* (5m/8m) = 5/8*v₀ ⇒v₀/vf = 8/5

4c) vf = 5/8*v₀

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the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

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Thus, the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

To obtain the change in the final speed of object B, apply the following equations.

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Thus, the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

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I understand that the question you are looking for is "Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistances of these two wires compare?

A) Both wires have the same resistance.

B) The longer wire has 1/3 the resistance of the shorter wire.

C) The longer wire has 3 times the resistance of the shorter wire.

D) The longer wire has 9 times times the resistance of the shorter wire.

E) The longer wire has 27 times times the resistance of the shorter wire."

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