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kenny6666 [7]
3 years ago
8

In a closed system, glider A with a mass of 0.40 kg and a speed of 2.00 m/s collides with glider B at rest with a mass of 1.20 k

g. The two interlock and move off. What speed are they moving at?
Physics
1 answer:
USPshnik [31]3 years ago
4 0
Conservation of momentum.
m1*u1 + m2*u2 =m1*v1 + m2*v2
here u2=0 body at rest, v1=v2=v both interlock and move together
0..4*2+1.2*0 = 0.4*v+1.2*v
v= 0.8/1.6 = 0.5 m/sec
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Answer: 80J

Explanation:

According to the first principle of thermodynamics:  

<em>"Energy is not created, nor destroyed, but it is conserved."  </em>

Then this priciple (also called Law) relates the work and the transferred heat exchanged in a system through the internal energy U, which is neither created nor destroyed, it is only transformed. So, in this especific case of the compressed gas:

\Delta U=Q+W  (1)

Where:

\Delta U is the variation in the internal (thermal) energy of the system (the value we want to find)

Q=-100J is the heat transferred out of the gas (that is why it is negative)

W is the work is done on the gas (as the gas is compressed, the work done on the gas must be considered positive )

On the other hand, the work done on the gas is given by:

W=-P \Delta V  (2)

Where:

P=450kPa=450(10)^{3}Pa is the constant pressure of the gas

\Delta V=V_{f}-V_{i} is the variation in volume of the gas

In this case the initial volume is V_{i}=600{cm}^{3}=600(10)^{-6}m^{3} and the final volume is V_{f}=200{cm}^{3}=200(10)^{-6}m^{3}.

This means:

\Delta V=200(10)^{-6}m^{3}-600(10)^{-6}m^{3}=-400(10)^{-6}m^{3}  (3)

Substituting (3) in (2):

W=-450(10)^{3}Pa(-400(10)^{-6}m^{3})  (4)

W=180J  (5)

Substituting (5) in (1):

\Delta U=-100J+180J  (6)

Finally:

\Delta U=80J  This is the change in thermal energy in the compression process.

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