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kramer
2 years ago
6

Two gliders are on a frictionless, level air track. Both gliders are free to move. Initially, glider A moves to the right and gl

ider B is at rest. After the collision, glider A has reversed direction and moves to the left. The mass of glider A is one quarter of the mass of glider B. The system is defined as the two gliders. Which object has the higher change in momentum?
Physics
1 answer:
Yuliya22 [10]2 years ago
4 0

Answer:

The change in momentum of both objects is the same but in opposite direction.

Explanation:

Hi there!

The momentum of the system is calculated as the sum of the momentums of each glider. The momentum of the system is conserved if no external force is acting on the objects (as in this case). That means that the initial momentum of the system is equal to the final momentum of the system.

The momentum of each glider is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass of the glider.

v = velocity.

The momentum of the system for glider A and B can be calculated as follows:

initial momentum = mA · vA + mB · vB

Where:

mA and vA = mass and velocity of glider A

mB and vB = mass and velocity of glider B

Initially, glider B is at rest so that vB = 0. Then, the initial momentum of the system is:

initial momentum = mA · vA

The final momentum of the system is calculated as follows:

final momentum = mA · vA´ + mB · vB´

Where vA´ and vB´ are the final velocities of glider A and B respectively.

We know that mB = 4mA and that vA´ is negative. The the final momentum will be:

final momentum = -mA · vA´ + 4mA · vB´

Since initial momentum = final momentum:

mA · vA = -mA · vA´ + 4mA · vB´

mA · vA + mA · vA´ = 4mA · vB´

<u>vA + vA´ = 4 vB´</u>

<u />

The change in momentum of glider A (ΔpA) is calculated as follows:

ΔpA = final momentum - initial momentum

ΔpA =  -mA · vA´ - mA · vA = -mA (vA + vA´) = -4mA · vB´

The change in momentum of glider B (ΔpB) is calculated as follows:

ΔpB = final momentum - initial momentum

ΔpB = 4mA · vB´ - 0 = 4mA · vB´

Then, the change in momentum of both objects is the same but in opposite direction. That´s why the momentum is conserved.

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gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle?  Assume the specific heat ratio is 1.2.

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T is the static temperature  = 2000 K

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Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

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Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa =  ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 =  Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

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Tt = ⁶√(4.3998976 × 10²¹)

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