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3 years ago

How is momentum conserved in a newtons cradle when one steel ball hits the other

Physics

1 answer:

arsen [322]3 years ago

8 0

In Newton's Cradle experiment we know that all cradles of same mass and identical to each other

Now we know that when two identical objects collide elastically then they interchange their velocity

So here we have same illustration

When Newton pulls up a cradle and release it will move hit another cradle which is at rest

Due to elastic collision between them first cradle comes to rest and another cradle will move ahead with same speed this process remains the same and one by one all cradle hit another.

So at the last the cradle at the end will move off with the same speed as the first cradle will hit with the speed.

So in this experiment the cradle at the last end will move off at same distance away from the right end as that of left end we pull the cradle.

So here we can say that in horizontal direction when all cradles are colliding each other there is no external force on the system so momentum is conserved and they all will move off with same speed and hence we observe the above condition.

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Suppose that the loudspeaker in the problem had a mass of 500 kg and the ropes hung 20∘ from the vertical. Into which of the fol

nexus9112 [7]

The following intervals would you expect T(tension) to fall : <u>2000 to 4000 N </u>

<h3>Further explanation</h3>

The force acting on a system with static equilibrium is 0

$\large{\boxed{\bold{\sum F=0}}$

(forces acting as translational motion only, not including rotational forces)

$\displaystyle \sum F_x = 0\\\\ \sum F_y = 0$

We complete the questions:

A 20-kg loudspeaker is suspended 2.0 m below the ceiling by two ropes that are each 30? from vertical.

Find the value of T, the magnitude of the tension in either of the ropes.

Express your answer in newtons.

Suppose that the loudspeaker in the problem has a mass of 500 kg and the ropes hung 20? from the vertical. Into which of the following intervals would you expect T to fall? You don't have to calculate Texactly to answer this question; just make an estimate.

500 to 1000 N

1000 to 2000 N

2000 to 4000 N

4000 to 8000 N

8000 to 16,000 N

In a 500-kg loudspeaker system and two ropes that are each 20° from vertical, we see the forces acting on the y axis (vertical)

$\displaystyle \sum F_y = 0\\\\T1_y+T2_y-w=0\\\\T1~cos~20^o+T2~cos~20^o=m\times g$

we assume g = 10 m/s², then :

$\displaystyle 2T~cos~20=500\times 10\\\\T~cos~20=2500\\\\\boxed{\bold{T=26592 N}}}$

So the value of T is between 2000 and 4000 N

<h3>Learn more</h3>

Newton's Law

brainly.com/question/13725525

Keywords : the loudspeaker, ropes, Tension, mass, intervals

#LearnWithBrainly

4 0

4 years ago

A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a

Gelneren [198K]

Answer:we can calculate the answer by following steps;

Explanation:

7 0

3 years ago

the star of Africa, a diamond in the royal scepter of the British crown jewels, has mass of 530.2 carats, where 1 carat = 0.20 g

QveST [7]

Given:
Mass of the diamond = 530.2 carats
1 carat = 0.20 g = 0.20 x 10⁻³ kg
1 kg = 2.21 lb

Therefore the mass of the diamond is
$(530.2 \, carats)*(0.20 \times 10^{-3} \, \frac{kg}{carat})*(2.21 \, \frac{lb}{kg}) = 0.2343 \, lb$

Answer: 0.2343 lb

6 0

3 years ago

A coal train in Montana starts from rest and accelerates at 2.5m/s^2 for 15 s.

bearhunter [10]

Complete Question

A coal train in Montana starts from rest and accelerates at 2.5m/s^2 for 15 Find the trains maximum velocity

Answer:

The maximum velocity is $v = 37.5 \ m/s$

Explanation:

From the question we are told that

The acceleration is $a =2.5\ m/s^2$

The time for the acceleration is $t = 15 \ s$

Generally from the kinematics equation

$v = u + at$

here u is the initial velocity and the value is 0 m/s given that it started rom rest

$v = 0 + 2.5 * 15$

$v = 37.5 \ m/s$

4 0

4 years ago

a 1200 kg car rolling on a horizontal surfact has a speed of 25 m/s when it strikes a horizontal coiled spring and is brought ot

gizmo_the_mogwai [7]

The required spring constant:

The spring constant of the spring is $12\times 10^4 \text{ N/m}$ .

Calculation:

The mass of the car is m=1200 kg, the speed of the car is v=25 m/s, and after colliding the spring is brought to rest at a distance of x=2.5m. Let the spring constant of the spring is, k.

From the conservation of energy,

Total initial kinetic energy= Total final potential energy of the spring

Therefore,

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

Now, substituting the values of the mass of the car, speed of the car, and displacement, we get:

$\begin{aligned}1200\times(25)^2&=k\times (2.5)^2\\\Rightarrow k&=\frac{1200\times(25)^2}{(2.5)^2}\\&=12\times 10^4 \text{ N/m}\end{aligned}$

To know more about spring constant, refer to:

brainly.com/question/14159361

#SPJ4

5 0

2 years ago