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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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Answer:

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In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 s later at height h ???? 20.0 m above the release level. The ball’s
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"Fig is attacted with answer"

Answer:

a) d = 33.72 m

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c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

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Angle = θ = 60°

Horizontal distance = d = ?

Using 2nd  equation of motion

h = v_{y_{f}}t + \frac{1}{2}gt^{2}

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v_{y_{f}}  = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

v_{f} = v_{y_{f}} / sinθ

v_{f} = 14.6 / sin 60

v_{f} = 16.86 m/s

v_{x_{f}} = v_{f}cosθ

v_{x_{f}} = 16.86 cos 60

v_{x_{f}} = 8.43 m/s

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d = v_{y_{f}} t

d = (8.43)(4)

d = 33.72 m

b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

v_{x_{f}} = v_{x_{i}}

but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.

So,

g = (v_{y_{f}} - v_{y_{i}} ) / t

9.8 =  14.6 - v_{y_{i}}) / 4

v_{y_{i}} = 24.6 m/s

v_{i} = \sqrt{v_{x_{i}}^{2}+v_{y_{i}}^{2} }

v_{i} = \sqrt{8.43^{2}+24.6^{2}}

v_{i} = 26 m/s

c)

cos β = v_{x_{i}} / v_{i}

β = cos⁻¹ (8.43 / 26)

β = 71.08°

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Answer:

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