1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Sladkaya [172]
3 years ago
13

Two balls undergo a perfectly elastic head-on collision, with one ball initially at rest. if the incoming ball has a speed of 20

0 m/s . part a part complete what is the final speed of the incoming ball if it is much more massive than the stationary ball? express your answer using two significant figures. v1 = 200 m/s submitprevious answers correct part b part complete what is the final direction of the incoming ball with respect to the initial direction if it is much more massive than the stationary ball? forward submitprevious answers correct part c part complete what is the final speed of the stationary ball if the incoming ball is much more massive than the stationary ball?.
Physics
2 answers:
melamori03 [73]3 years ago
7 0
what is the final speed of the incoming ball if it is much more massive than the stationary ball? express your answer using two significant figures. v1 = 200 m / s submitprevious answers correct
 Perfectly elastic collisions means that both mechanical energy and
momentum are conserved.
 Therefore, for this case, we have the equation to find the final velocity of the incoming ball is given by
 v1f = ((m1-m2) / (m1 + m2)) v1i
 where,
 v1i: initial speed of ball 1.
 v1f: final speed of ball 1.
 m1: mass of the ball 1
 m2: mass of the ball 2
 Since the mass of the ball 1 is much larger than the mass of the ball 2 m1 >> m2, then rewriting the equation:
 v1f = ((m1) / (m1) v1i
 v1f = v1i
 v1f = 200 m / s
 answer
 200 m / s
 part b part complete what is the final direction of the incoming ball with respect to the initial direction if it is much more massive than the stationary ball? forward submitprevious answers correct

 Using the equation of part a, we can include in it the directions:
 v1fx = ((m1-m2) / (m1 + m2)) v1ix
 v1i: initial velocity of ball 1 in the direction of the x-axis
 v1f: final speed of ball 1 in the direction of the x-axis
 like m1 >> m2 then
 v1fx = v1ix
 v1fx = 200 m / s (positive x direction)
 So it is concluded that the ball 1 continues forward.
 answer:
 forward


 part c part complete what is the final speed of the stationary ball if the incoming ball is much more massive than the stationary ball ?.
 The shock is perfectly elastic. For this case, we have that the equation to find the final velocity of the stationary ball is given by
 v2f = ((2m1) / (m1 + m2)) v1i
 where,
 v1i: initial speed of ball 1.
 v2f: final speed of ball 2.
 m1: mass of the ball 1
 m2: mass of the ball 2
 Then, as we know that m1 >> m2 then
 v2f = ((2m1) / (m1) v1i
 v2f = 2 * v1i
 v2f = 2 * (200 m / s)
 v2f = 400 m / s
 answer
 400m / s
weqwewe [10]3 years ago
4 0

a) The final speed of the massive incoming ball is \boxed{200\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}.

b) <u>The direction of the motion of the incoming ball with respect to its initial position is in the forward direction</u>.<u> </u>

c) The final speed of the stationary ball due to the incoming massive ball is \boxed{{\text{400}}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}.

Further Explanation:

Given:

The speed of the incoming ball is 200\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}.

The second ball is initially at rest So, its initial velocity is 0\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}.

Concept:

<u>Part (a): </u>

The two bodies moving towards one another when collide elastically; the momentum as well as the mechanical energy of the system remains conserved at the time of collision.

By using the conservation of momentum at the time of collision:

Mu{ & _1}+m{u_2}=M{v_1}+m{v_2}                                …… (1)

Here, M is the mass of the massive ball, m is the mass of lighter ball, {u_1}\,\& \,{u_2} are the initial speeds of the two bodies and {v_1}\,\& \,{v_2} are the final speeds of the bodies.

Substitute the values of {u_1}\,\& \,{u_2} in above expression.

M\times200+m\times0=\left({M\times {v_1}} \right)+\left({m\times{v_2}}\right)

SinceM>> >m, the above expression can be rearranged as:

\begin{aligned}200\times M&=\left(M\right){v_1}\\{v_1}&=200\left({\frac{M}{M}} \right)\\&=200\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}

Thus, the final speed of the incoming ball at collision is \boxed{200\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}.

<u>Part (b): </u>

The ball is moving in the forward direction and has no other velocity other than the forward direction. Initially, the ball started in the forward direction and finally it is moving with the speed 200\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} in the forward direction only.

Therefore, it suggests that the massive ball is moving in forward direction with respect to its initial direction.

<u>Part (c): </u>

In order to obtain the final speed of the smaller ball after the collision, rearrange the equation (1) to obtain the final velocity of the second ball {v_2}.

\begin{aligned}{v_2}&={v_1}\left({\frac{{2M}}{M}}\right)\\&=200\left(2\right)\\&=400\,{{\text{m}} \mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}

Thus, the final speed of the smaller ball after collision with the massive ball is \boxed{400\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}} in forward direction.

Learn More:

1. A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s brainly.com/question/9484203

2. What is the magnitude of the force that the car exerts on the truck brainly.com/question/2235246

3. A car traveling in a straight line at an initial speed of 8.0 meters per second accelerates uniformly brainly.com/question/6423792

Answer Details:

Grade: College

Subject: Physics

Chapter: Momentum Conservation

Keywords:

Momentum conservation, massive ball, smaller ball, perfectly elastic head-on collision, incoming ball has speed, final speed of the incoming ball.

You might be interested in
I dont know what pfp to put on today so I'm asking u guys....
Mandarinka [93]

Answer:

i think number 2 should be your pfp

5 0
3 years ago
Which is the result of using a machine?
Eddi Din [679]
Correct Answers is A.

The machines gives us some mechanical advantage. This means the mechanical average makes the work output greater than the work input
Simple most example is a lever. The force applied is smaller and the output work is larger as compared to input.

Option B cannot be true, as there must be a force to get some work done.
Option C and D are inverse of what a machine is designed for. A small force can be exerted through a large distance to have a large force exerted through a small distance. Common Example of this principle is a screw opener. 
3 0
3 years ago
8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer
Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.​

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \  by \ the \ effort}{disatnce \ moved \  by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m

Therefore, the distance that the object is raised above its initial position is 5.625 m.​

3 0
3 years ago
A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.070
likoan [24]

Given Information:

length of slender rod = L = 90 cm = 0.90 m

mass of slender rod = m = 0.120 kg

mass of sphere welded to one end = m₁ = 0.0200 kg

mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)

Required Information:

Linear speed of the 0.0500 kg sphere = v = ?

Answer:

Linear speed of the 0.0500 kg sphere = 1.55 m/s

Explanation:

The velocity of the sphere can by calculated using

ΔKE = ½Iω²

Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy

The moment of inertia of a rigid rod about center is given by

I = (1/12)mL²

The moment of inertia due to m₁ and m₂ is

I = (m₁+m₂)(L/2)²

L/2 means that the spheres are welded at both ends of slender rod whose length is L.

The overall moment of inertia becomes

I = (1/12)mL² + (m₁+m₂)(L/2)²

I = (1/12)0.120*(0.90)² + (0.0200+0.0500)(0.90/2)²

I = 0.0081 + 0.01417

I = 0.02227 kg.m²

The change in the potential energy is given by

ΔPE = m₁gh₁ + m₂gh₂

Where h₁ and h₂ are half of the length of slender rod

L/2 = 0.90/2 = 0.45 m

ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45

The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.

ΔPE = -0.1323 J

This potential energy is then converted into kinetic energy therefore,

ΔKE = ½Iω²

0.1323 = ½(0.02227)ω²

ω² = (2*0.1323)/0.02227

ω = √(2*0.1323)/0.02227

ω = 3.45 rad/s

The linear speed is

v = (L/2)ω

v = (0.90/2)*3.45

v = 1.55 m/s

Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.

8 0
3 years ago
A cart with a mass of 0.5 kg is at the top of the ramp. The height is 0.40m .
Tju [1.3M]

A=0.05.0M.

B=68.9244GPE.34

C=0

D it would be 79%HIGHER

3 0
3 years ago
Other questions:
  • Alcohol occurs when the desire to feel good and have the pleasurable feelings associated with alcohol or other drugs drives desi
    11·1 answer
  • A force or an event that changes the earth's surface is known as
    11·1 answer
  • If P = 4.00î + 3.00k, P · Q = 17.0, and P ✕ Q = −6.00ĵ, determine the vector Q.
    11·1 answer
  • What was the name of the first steam-powered locomotive that was developed by george stephenson?
    7·1 answer
  • A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
    7·1 answer
  • Write down the role played by four digestive juices in digestion and the
    10·1 answer
  • 11. (a)What downward force is acting on you when you go down a waterslide? (b)What type of friction is
    11·1 answer
  • Click on image to see it <br><br> please show all steps and formulas
    5·2 answers
  • Find the direction of this
    6·1 answer
  • the ability of torricelli's hypothesis about atmospheric pressure to suggest the design of the barometer is an illustration of:
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!