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HACTEHA [7]
3 years ago
15

A 0.17kg ball rolls at 0.75m/s to the right on a frictionless surface and collides with a 0.17kg ball rolling to the left at 0.6

5m/s. (Elastic Collision) What is the final velocity of the second ball when the first is 0.5m/s to the left? What is the change in kinetic energy before and after?
Physics
1 answer:
mario62 [17]3 years ago
8 0

Both momentum and kinetic energy are conserved in elastic collisions (assuming that this collision is perfectly elastic, meaning no net loss in kinetic energy)

To find the final velocity of the second ball you have to use the conversation of momentum:

*i is initial and f is final*

Δpi = Δpf

So the mass and velocity of each of the balls before and after the collision must be equal so

Let one ball be ball 1 and the other be ball 2

m₁ = 0.17kg

v₁i = 0.75 m/s

m₂ = 0.17kg

v₂i = 0.65 m/s

v₂f = 0.5

m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f

Since the mass of the balls are the same we can factor it out and get rid of the numbers below it so....

m(v₁i + v₂i) = m(v₁f + v₂f)

The masses now cancel because we factored them out on both sides so if we divide mass over to another side the value will cancel out so....

v₁i + v₂i = v₁f + v₂f

Now we want the final velocity of the second ball so we need v₂f

so...

(v₁i + v₂i) - v₁f = v₂f

Plug in the numbers now:

(0.75 + 0.65) - 0.5 = v₂f

v₂f = 0.9 m/s


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A vertical scale on a spring balance reads from 0 to 250 N. The scale has a length of 15.0 cm from the 0 to 250 N reading. A fis
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Frequency f=2.60 Hz

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Mass of fish m

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k=Force/distance\\k=F/x\\k=250/0.15\\k=1666.7 N/m

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5. A stream of air at 12 bar and 900 K is mixed with another stream of air at 2 bar and 400 K with 2 times the mass flowrate. If
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Answer:

The anserrs to the question are

(a)  The temperature would be 566.67  K and

(b) The pressure of the resulting air stream is 14 bar

Explanation:

Here the two streams of gases meet ad form a single stream

The steady-flow energy equation can be implemented at the mixing point of the to streams as follows

mₐhₐ₁ + mₙ hₙ₁ + Q° + W° = mₐhₐ₂ + mₙhₙ₂

Where the flow is adiabatic, we have

Q = 0 and  W = 0  hence

mₐhₐ₁ + mₙ hₙ₁ = mₐhₐ₂ + mₙhₙ₂ where h = cp×T we have

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Therefore the final temperature T is given by

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Therefore we have where Tₐ = = 900 K and Tₙ = 400 K and mₙ = 2mₐ gives

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From Dalton's law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the components of the mixture

Therefore the total pressure of the combined stream = pₐ + pₙ = p

= 12 bar + 2 bar = 14 bar

Stream pressure = 14 bar

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