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Rom4ik [11]
3 years ago
6

A ball collides with a vertical, unmovable wall. There is no friction between the wall and the ball (the only force acting on th

e ball during the collision is the normal force from the wall). ball colliding with wall 1) What is the direction of the x-component of the impulse imparted to the ball during the collision?
Physics
1 answer:
user100 [1]3 years ago
7 0

Answer: the same direction I.e to the left.

Explanation:

The component perpendicular to the contact surface is such that will stop the relative motion and, in case of elastic collision like here, return the system to the same kinetic energy. So ball hitting immovable surface will have the same speed (magnitude of velocity) as before the collision.

There will also be parallel force caused by friction, but it has to be treated separately for two reasons:

The perpendicular force is limited to coefficient of friction times the normal force. If that is not enough to stop the ball, it will skid on the surface.The perpendicular force, and this depends on the specific geometry, does not pass through the centre of mass of the ball. Therefore it imparts a moment on the ball that causes it to start rotating. And once the ball is rotating so that the point of contact is stationary, there is no momentum to cause any friction force anymore and the friction force disappears and stops decelerating the ball.

So what happens is that the vertical component of the velocity will be reversed, while the horizontal component will be somewhat reduced with the corresponding amount of kinetic energy transferred to energy of rotation. The rotation will always eliminate the friction force before the horizontal component of velocity is zeroed, so the ball will always continue in the same direction, just a bit slower.

If you instead threw an elastic box (which could not start rotating freely) it could actually bounce back.

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Leokris [45]
Newtons 3.law: Action = Reaction

If a body exerts a force on a rope of 400 N the rope exerts a force on the body of 400N also. So the tension in the rope is 400N. See pictures below.

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4 years ago
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ohaa [14]

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3 years ago
You place a light bulb 8 cm in front of a concave mirror. You then move a sheet of paper back and forth in front of the mirror.
Alika [10]

sorry - late reply...just stumbled across tis...hope u can still use it :)


By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
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5 0
4 years ago
a water bomber flying with a horizontal speed of 85m/s at a height of 3000m drops a load on a fire below. How far in front of th
Andreyy89

Answer:

2081.65 m

Explanation:

We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:

Height (h) = 3000 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

3000 = ½ × 10 × t²

3000 = 5 × t²

Divide both side by 5

t² = 3000 / 5

t² = 600

Take the square root of both side

t = √600

t = 24.49 s

Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:

Horizontal velocity (u) = 85 m/s

Time (t) = 24.49 s

Horizontal distance (s) =?

s = ut

s = 85 × 24.49

s = 2081.65 m

Thus, the load should be released from 2081.65 m.

3 0
3 years ago
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