Collision Lab
This activity will help you meet these educational goals:

You will explain or predict phenomena by exploring qualitative relationships between variables.
You will use positive and negative numbers to represent quantities in real-world contexts.
Directions
Read the instructions for this self-checked activity. Type in your response to each question, and check your answers. At the end of the activity, write a brief evaluation of your work.
Activity
Open this collision simulator and click Introduction. You’ll use the simulator to explore and compare elastic collisions and inelastic collisions. The mass and starting velocity of the colliding objects are kept constant. Follow the instructions in each part, and then answer the questions that follow. Use the math review if you need help with adding and subtracting negative numbers.

Question 1: Elastic Collisions
In this question, you will investigate elastic (bouncy) collisions. Be sure that the slider is to the extreme right (elasticity 100%).

Part A
Click Show Values in the upper-right corner. Study the boxes on the screen. What are the mass and initial velocity of ball 1 and ball 2?

I NEED HELP!


Part B
Part B
Click Play, and watch the balls collide. Then click Pause. What are the final velocities of ball 1 and ball 2?


The number line shows the starting and ending velocities for ball 1. What’s the change in velocity of ball 1? Calculate the value mathematically, and check it using the number line.

a number line showing an ending velocity of -0.50 meter/second and a starting velocity of 1.00 meter/second

Answers

Answer 1

Answer:

Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

Explanation:

Answer 2

Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

What is Collision?

A collision is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.

The following are a few instances of physical encounters that scientists might classify as collisions. Legs of an insect are said to collide with a leaf when it falls on one.

Every contact of a cat's paws with the ground while it strides across a lawn is seen as a collision, as is every brush of its fur with a blade of grass.

Therefore, Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

To learn more about collision, refer to the link:

https://brainly.com/question/13138178

#SPJ2


Related Questions

What is the acceleration of the the object during the first 4 seconds?

Answers

Answer:

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Explanation:

What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

convert 100 Newton into dyne​

Answers

Answer:10000000

Explanation:

It would actually be 10 million dyne

A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m. The force of friction between the refrigerator and the floor is 230 N. How much work has been performed by the mover on the refrigerator?​

Answers

Given :

A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m.

The force of friction between the refrigerator and the floor is 230 N.

To Find :

How much work has been performed by the mover on the refrigerator.

Solution :

Since, refrigerator is moving with constant velocity.

So, force applied by the mover is also 230 N ( equal to force of friction ).

Now, work done in order to move the refrigerator is  :

[tex]W = Force\times distance\\\\W = 230 \times 8.1\ N\ m\\\\W = 1863\ N\ m[/tex]

Hence, this is the required solution.

The fact that our preconceived ideas contribute to our ability to process new information best illustrates the importance of: the serial position effect. O repression iconic memory . semantic encoding . retroactive interference .

Answers

Answer:

It’s a

Explanation:

Don’t actually put that i needed the points mb

Galaxy B moves away from galaxy A at 0.577 times the speed of light. Galaxy C moves away from galaxy B in the same direction at 0.731 times the speed of light. How fast does galaxy C recede from galaxy A?

Answers

Answer:

The value is  [tex]p = 0.7556 c[/tex]

Explanation:

From the question we are told that

   The speed at which galaxy B moves away from galaxy A is  [tex]v = 0.577c[/tex]

Here c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

     The speed at which galaxy C moves away from galaxy B is  [tex]u = 0.731 c[/tex]

Generally from the equation of  relative speed we have that  

     [tex]u = \frac{p - v}{ 1 - \frac{ p * v}{c^2} }[/tex]

Here p is the velocity at which galaxy C recede from galaxy A so

     [tex]0.731c = \frac{p - 0.577c }{ 1 - \frac{ p * 0.577c}{c^2} }[/tex]

=>   [tex]0.731c [1 - \frac{ p * 0.577}{c}] = p - 0.577c[/tex]

=>   [tex]0.731c - 0.4218 p = p - 0.577c[/tex]

=>   [tex]0.731c + 0.577c = p + 0.4218 p[/tex]

=>   [tex]1.308 c = 1.731 p[/tex]

=>    [tex]p = 0.7556 c[/tex]

A golf ball flies through the air after being struck with a golf club. Which of the following statements describes the force on the ball as momentum is transferred between the club and ball?

A. The ball does not experience any force.
B. The force experienced by the ball is greater than the force experienced by the club.
C. The force experienced by the ball equals the force experienced by the club.
D. The force experienced by the ball is weaker than the force experienced by the club.

Answers

Answer:

C) The force experienced by the ball equals the force experienced by the club.

Explanation:

When the golfer strikes the ball with his club, the club exerts a force on the ball. Due to Newton's Third Law Of Motion [Every Action has an equal and opposite reaction], the ball also exerts an equal force on the club. However,

As the mass of the club is usually greater than the mass of the ball, it accelerates slower; While the ball way faster, following the equation : F=ma

Which objects cannot be observed in detail without a microscope?

Answers

Answer:

partecls

Explanation:

because they are to small to see with plain eyes

A projectile is fired horizontally from a height of 10 m above level ground. The projectile lands a horizontal distance of 15 m from where it was launched.
-Find the hang time for the projectile.
-Find the initial speed of a projectile.
-What are the x and y components of the projectile’s velocity the moment before it strikes the ground?
-At what speed will the projectile strike the ground?

Answers

Answer:

a)t  = 1,43 s

b) V = 10,49 m/s

c) V₀ₓ = 10,49 m/s   ;    V₀y = 14,01 m/s

d) Vf = 17,5 m/s

Explanation:

According to the problem statement

V₀ = V₀ₓ    and  V₀y = 0

And at the end of the movement t = ?  the distance y = 10 m

Therefore as

h = V₀y - (1/2)*g*t²

Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²

10 = 4,9*t²

t² = 10/4,9    ⇒  t² = 2,04 s

t  = 1,43 s

a) 1,43 s is the time of movement

b) V₀ = V₀ₓ        V₀y = 0     and  V₀ₓ = Vₓ     ( constant )

Just before touching the ground, the horizontal distance is

hd = 15 = Vₓ * t

Then  15 /1,43 = Vₓ = V₀ₓ

Vₓ = 10,49 m/s

Then initial speed is V = 10,49 m/s    since V₀y = 0

Vf² = Vₓ² + Vy²

Vyf = V₀y - g*t

Vyf =  0 - 9,8 *1,43

Vyf = - 14,01 m/s

And finally the speed when the projectile strike the ground is:

Vf² = Vₓ² + Vy²

Vf = √ (10,49)² + (14,01)²

Vf = 17,50 m/s

How was the Periodic Table of Elements developed and how are the elements arranged on it?

Answers

Answer:

In 1869 Russian chemist Dimitri Mendeleev started the development of the periodic table, arranging chemical elements by atomic mass. He predicted the discovery of other elements, and left spaces open in his periodic table for them.

Explanation:

Answer: Mendeleev first published a table of elements arranged according to increasing atomic masses. He noticed that some elements near each other had differing properties, but elements in vertical columns had similar properties. Moseley then rearranged the table according to atomic numbers and this eliminated the discrepancies found in Mendeleev’s attempt. Today’s version of the periodic table displays elements in order based on their atomic number; the atomic number indicates the number of protons within the atoms of a particular element. Rows are called periods and columns are called groups. Elements in the same group have similar properties. Elements are grouped into nine categories: noble gases, halogens, nonmetals, alkali metals, alkaline earth metals, transition metals, other metals, metalloids, and rare earth elements.

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