3.the collision.The conservation of the total momentum before and after the collision is expressed by:m2 2 lu = in?u + In luLikewise, the conservation of the total kinetic energy is expressed by:Eatru + a lu = ntu + luThese equations may be solved directly to find v1, v2 when u1, u2 are known:Tu - lu+ InZu + Tuu2mị + m22m1+In-Tu + Tu7n-u + Tuv2(1) 2.41 m/s south, 7.39 m/s north25.30 m/s south, 2.05 m/s north4.85 m/s north, 3.20 m/s south45.10 m/s north, 2.89 m/s south A bumper car of mass 125 kg, moving at 6.30 m/s to the north, collides 100% elastically, head-on, with a second bumpercar of mass 100 kg moving at 3.50 m/s to the south. What are the velocities of the first and second bumper carsrespectively?Note: the following may come in handy here! In these formulas, uj and uz are the velocities of the two cars before thecollision, and v and v, are the velocities after the collision.VIThe conservation of the total momentum before and after the collision is expressed by:m1u1 + m2 U2 = m1v1 + m2v2.Likewise, the conservation of the total kinetic energy is expressed by:m už +mzu = mvf +m.These equations may be solved directly to find v1, v2 when u1, u2 are known:2m2m2U1 +mị + m2mịmị + m22m1m2 – mị-U1 +mị + m2V2m1 + m21-2.41 m/s south, 7.39 m/s northSave for Later

Given that, Mass of first bumper(m1)= 125 kg Mass of second bumper (m2)= 100 kg Velocity of first…

the collision. The conservation of the total momentum before and after the collision is expressed by: m1u1 + m2U2 = m1vi + m2v2. Likewise, the conservation of the total kinetic energy is expressed by: Entru + fa tu = ntru? + n lu Entru + ļa tu? These equations may be solved directly to find v1, v2 when u1, uz are known: - + In. Zu + Tu m2 Tu + Tu · 2m1 + In- Tu + Tu v2 Tu + Tu (1 2.41 m/s south, 7.39 m/s north 2 5.30 m/s south, 2.05 m/s north 3 4.85 m/s north, 3.20 m/s south 4. 5.10 m/s north, 2.89 m/s south

the collision. The conservation of the total momentum before and after the collision is expressed by: m1u1 + m2U2 = m1vi + m2v2. Likewise, the conservation of the total kinetic energy is expressed by: Entru + fa tu = ntru? + n lu Entru + ļa tu? These equations may be solved directly to find v1, v2 when u1, uz are known: - + In. Zu + Tu m2 Tu + Tu · 2m1 + In- Tu + Tu v2 Tu + Tu (1 2.41 m/s south, 7.39 m/s north 2 5.30 m/s south, 2.05 m/s north 3 4.85 m/s north, 3.20 m/s south 4. 5.10 m/s north, 2.89 m/s south

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: A ball of mass 0.568 kg moving east (+x direction) with a speed of 3.90 m/s collides head-on with a…

A: Given:- Assume that the collision is perfectly elastic. A ball of mass M = 0.568 kg moving east…

Q: stationary 6.0 kg object blows apart into 3 pieces. The 3 pieces move away from each other on a…

A: mass of stationary object (M) = 6 kg mass of first piece (m1) = 2 kgmass of second piece (m2) = 1 kg…

Q: 1. A 7kg ball moving at 1m/s to the right collides with a 2kg ball which is moving at 8m/s to the…

A: Let the direction to right be positive, Total momentum before collision pi=7×1-2×8=7-16=-11 kg-m/s…

Q: A 48 g marble moving at 2.3 m/s strikes a 29 g marble at rest. Assume the collision is perfectly…

Q: 4. A bullet ( v = 60 m/sec, m = 50 gram) is fired into an ice block (M=3 KG) resting on ice…

A: Given information:Mass of the bullet (m) = 50 g = 0.05 kgVelocity of the bullet (u1) = 60 m/sMass…

Q: Two balls undergo a perfectly elastic head-on collision, with one ball initially at rest. If the…

A: Given data The stationary ball's initial velocity is: u1 = 0 m/s The initial velocity of the…

Q: An object has a kinetic energy of 275 J and a linear momentum of magnitude 25kg m/s. Find the mass…

Q: 1. Consider a bomb of mass M (see figure below) that sits in the middle of space. Initially, the…

A: Given, Mass of bomb, M masses of 4 portions of the bomb after the explosion will be, m1,m2,m3 and…

Q: You have two vectors: A = 7.27î + 3ĵ and B = = 4î — 0.95ĵ + 12k. If vector A represents an initial…

A: The initial momentum vector is and the final momentum vector is .

Q: QUESTION 1 Impulse O A force exerted over a time interval that causes a change in an object's…

Q: a) What would be the speed of the combination after the collision? b) How much did the spring…

Q: i. Draw the balls and the approximate momentum vectors before and after the collision. ii. What is…

A: Given: V1 =3.0 i^ m/sV2 =2.0 j^ m/sV1' =0.5 i^+1.5 j ^m/sm1=m2 =m

Q: Particle one of mass m initially moves to the right at v. Particle two of mass 2m is initially at…

A: The initial momentum of a particle of mass m is P1i⇀ The initial momentum of a particle of mass 2m…

Q: Yanick and Nancy drive two identical 1380 kilogram cars at 35 m/sec. Yanick slams on the brakes and…

Q: 4. A container explodes and breaks into three fragments that fly off in one plane, in directions…

Q: A superball of mass 0.25 kg is dropped from rest from a height of h= 1.50 m above the floor. It…

Q: One billiard ball is shot east at 2.4 m/s . A second, identical billiard ball is shot west at 0.80…

Q: A particle with kinetic energy equal to 244 J has a momentum of magnitude 21.7 kg · m/s. Calculate…

Q: Is this situation possible? If not, why not? A stationary particle fragments into two particles. One…

A: This situation is not possible.

Q: 5. Suppose an object is moving in the air. It has a kinetic energy of 273 J and a momentum of 25.2…

Q: swer the following questions. Note that e = 0.43. a. What is the velocity (m/s) of A after the…

A: Given:- The mass of the car A is M = 1.3 metric tons = 1300 kg the mass of the car B is m = 1.7…

Q: A student attending an online 1301 class in the Fall20 semester decided to check what would happen…

Q: 4. Two astronauts with masses mj and m2 are moving with the velocity u together. They push each…

Q: A snowball is thrown with a speed of 49.5 km/h, giving it momentum with a magnitude of 2.90 Ns. The…

A: The speed of the snowball is given by v = 49.5 km/h = 13.75 m/s The momentum p = 2.9 Ns = 2.9 kgm/s…

Q: A projectile is launched with 400 kg m/s of momentum and 2000 J of kinetic energy. What is the mass…

A: Given data: The projectile is p=400 kg.m/s The kinetic energy is K=2000 J=2000 kgm2/s2

Q: 1. a) For a particle of mass m and velocity v, what are the momentum, and its (kinetic) energy?…

Question

3.
the collision.
The conservation of the total momentum before and after the collision is expressed by:
m2 2 lu = in?u + In lu
Likewise, the conservation of the total kinetic energy is expressed by:
Eatru + a lu = ntu + lu
These equations may be solved directly to find v1, v2 when u1, u2 are known:
Tu - lu
+ In
Zu + Tu
u2
mị + m2
2m1
+In-
Tu + Tu
7n-
u + Tu
v2
(1) 2.41 m/s south, 7.39 m/s north
2
5.30 m/s south, 2.05 m/s north
4.85 m/s north, 3.20 m/s south
4
5.10 m/s north, 2.89 m/s south

A bumper car of mass 125 kg, moving at 6.30 m/s to the north, collides 100% elastically, head-on, with a second bumper
car of mass 100 kg moving at 3.50 m/s to the south. What are the velocities of the first and second bumper cars
respectively?
Note: the following may come in handy here! In these formulas, uj and uz are the velocities of the two cars before the
collision, and v and v, are the velocities after the collision.
VI
The conservation of the total momentum before and after the collision is expressed by:
m1u1 + m2 U2 = m1v1 + m2v2.
Likewise, the conservation of the total kinetic energy is expressed by:
m už +mzu = mvf +m.
These equations may be solved directly to find v1, v2 when u1, u2 are known:
2m2
m2
U1 +
mị + m2
mị
mị + m2
2m1
m2 – mị
-
U1 +
mị + m2
V2
m1 + m2
1-
2.41 m/s south, 7.39 m/s north
Save for Later

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

A block of mass m=17 kg starts moving 100 ms^-1 up a frictionless hill of height 5 m. It hits a stationary block mass M=4kg at the top of the hill. They stick together and move away along a level, frictionless surface. What’s their velocity after collision?
So far we have learned that the formula for Kinetic Energy is KE= 1/2mv2. However, for this lab, we would like to describe the motion of the objects using momentum. p=mv, instead of velocity. Using the two equations given, find a new formula for Kinetic Energy in terms of momentum not velocity. Based on the picture, which one is the correct answer?
A spring with relaxed length 0.30 m and spring constant 1.8 N/m rests horizontally on a frictionless table and is anchored frmly to a wall on one side. You press a 400 g block into the other side until the spring length is 0.16 m and then let go. a.) What will the momentum of the block be when it departs from the spring? (Assume contact ceases when the spring is relaxed.) b.) If the table were no longer frictionless and the block-table interface had a kinetic coeffcient of friction of 0.1 what then would the momentum of the block be when it departs from the spring?
this is a multi step problem but I am just having trouble with this part of the problem. Here are the values that I correctly calculated: mass of particle 1=6 kg initial velocity for particle 1= 185i+138j final velocity for particle 1= -68.89i+-50.67j I am trying to find: What is the change in kinetic energy for particle 1? I understand I have to use KT= 1/2*m*(vf2-vi2) but I keep getting the wrong answer can you please help?
1. Let's say you are measuring the velocity of an object before and after a collision. Let the mean velocity before the collision be v, and the mean velocity after the collision If the errors in both velocities are given by , and , respectively, then what is the error in the difference between the two velocities given by v, - v, ? be V2 2.
A .97 kg projectile moves at 19 m/s a. What is its momentum? b. How fast would a 5kg ball have to move to have the same momentum?
Railroad cars collide and stick together. How fast are the moving after sticking? Railroad car A is 1200kg and moving at 20.0m/s. Railroad car B is 8000kg and is not moving.
how to do?
A small spaceship whose mass is 2200 kg (including an astronaut) is drifting in outer space with negligible gravitational forces acting on it. If the astronaut turns on a 12 kW laser beam, what speed will the ship attain in 6.0 days because of the momentum carried away by the beam? Number i Units
If momentum is conserved in a collision, does this indicate conclusively that the collision is elastic? Explain. A.Yes, momentum is conserved only in an elastic collision. B. No, momentum is conserved only in inelastic collisions. C No, momentum is conserved in all collisions, elastic or inelastic.
m Р V Calculating Momentum, Mass, and Velocity 0 Directions: Solve the following problems by entering values into the red boxes. Stop decimals at the tenths place (use 1 decimal). When all entries are GREEN, you are done. Don't use spaces when entering values! Question 1 A bicycle has a momentum of 125 kg*m/s and a velocity of 2.0 m/s. What is the bicycle's mass? Question 2 A baseball has a momentum of 21 kg"m/s. What the baseball's mass if it has a velocity 7.0 meters per second to cover the distance? Formula to use: Question 3 What is the velocity of an emu that has a momentum of 1800 kg*m/s and a mass of 45 kg? Question 4 What is the momentum of a 1500 kg car that has a velocity of 4.0 m/s? Formula to use: Question 6 What is the momentum of a 6.5 kg bowling ball rolled down an alley with a velocity of 3.0 m/s? Question 5 A linebacker has a momentum of 60 kg*m/s and a velocity of 3.0 m/s. What is the linebacker's mass? Formula to use: Formula to use: Question 9 Formula to use:…
My last step was F=.57mg/Cos(-15.1)+(0.57 Sin (15.1) Outcome of 0.6977mg I think I found the The condition for moving a block but I don’t know what to do next.I’m confused