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fredd [130]
3 years ago
14

In midair in the international space station a 1 kg chunk of putty moving at 1 m/s collides with and sticks to a 5 kg chunk of p

utty initially at rest. at what speed does the 6 kg chunk travel after the collision?
Physics
1 answer:
Delicious77 [7]3 years ago
6 0
<span>We know that the momentum keeps constant in a inelastic collisions, so the product of mass and speed do not change:
   m1 * v1 + m2 * v2 = m * v
 1 * 1 + 5 * 0 = (1 + 5) * v
  1 = 6 * v
 v = 1/6 m/s
   So the final speed of the 6 kg chunk will travel at 0.167 m/s</span>
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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
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Answer:

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

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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

3 0
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Explanation:

Data provided in the problem:

The formula for conversion as:

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Now,

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Substituting the value of F in the above formula, we get

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