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Karo-lina-s [1.5K]
2 years ago
14

How does Kinetic Energy and the Law of Conservation of Momentum apply to inelastic collisions?

Physics
2 answers:
Nadya [2.5K]2 years ago
7 0

Answer:

Explanation:

Why does the inelastic collision momentum remain conserved while the kinetic energy changes?

Imagine two objects colliding, M1 and M2. Let’s say that M1 slows down and M2 speeds up. The change in momentum for M1 will be equal to the Impulse acting on M1 which is defined as the force on M1 multiplied by the time interval the force acts. The change in momentum for M2 will be the force on M2 multiplied by that SAME time interval. The two forces are an action-reaction pair; M2 pushing on M1 and M1 pushing back on M2.

Since the two forces are equal and opposite and the time intervals are equal, the two Impulses must also be equal and opposite, and so will cancel each other out. If the TOTAL Impulse is zero there will be NO change in total momentum. So momentum must be conserved during the collision.

The situation with kinetic energy is different. Kinetic energy is changed by Work, defined as force times distance. While the two forces acting on the two objects are equal and opposite (as described above), the distances traveled by the two objects during the collision are NOT necessarily equal to each other. So kinetic energy is not automatically conserved.

One of the requirements of an inelastic collision is one in which the two objects colliding do NOT travel the same distance during the collision.

Lorico [155]2 years ago
5 0

Answer:

Explanation:

An inelastic collision is one in which the internal kinetic energy changes (it is not conserved) . The two objects come to rest after sticking together, conserving momentum. But the internal kinetic energy is zero after the collision.

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A crowbar having the length of 1.75m is used to balance a load between the fulcrum and the load is 0.5m calculate MA
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Explanation:

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3 0
3 years ago
A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.5 cm before reaching its new equilibrium
Brums [2.3K]

Answer:

0.99Hz

Explanation:

Using F= -mx ( spring force)

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A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past t
marta [7]

Answer:

Explanation:

Given that

The window height is 2m

And the window is 7.5m from the ground

Then the total height of the window from the ground is 7.5+2=9.5m

It takes the ball 0.32sec travelled pass the window.

When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity ( v')

Now using the equation of free fall during this window travels

S=ut-½gt² against motion.

S=2, g=9.81, t=0.32sec

Then,

S=u't-½gt²

2=u'×0.32-½×9.81×0.32²

2=0.32u'-0.5023

2+0.5032=0.32u'

Then, 0.32u'=2.5032

u'=2.5032/0.32

u'=7.82m/s

This is the initial velocity as the ball got the the window

Now, let analyse from the window bottom to the ground which is a distance of 7.5m

Using the equation of free fall again

v²=u²-2gH

In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i.e u'=7.82m/s,

While u is the original initial velocity from the throw of the ball

Then,

u'²=u²-2gH

7.82²=u²-2×9.81×7.5

61.146=u²-147.15

61.146+147.15=u²

Then, u²=208.296

So, u=√208.296

u=14.43m/s

The initial velocity of the ball form the throw is 14.43m/s

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Answer:

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