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

Similar questions

The first moving object is moving west at 10 m/s. The second moving object is moving west at 15 m/s. What is the momentum of the system before and after the collision? What is the velocity of the moving objects after the collision? My teacher said the collision is sticking, but how do we know that? Collisions are elastic. The first moving object and the second moving object both weigh 1500 kg. Show unknowns and knowns Use formula mv (before) = mv (after) and then solve for v (after)
Help me please. Which is the correct letter/choice to number 4 and 6?
Must solve the question completely. Don't solve partially. Skip it pls if u don't want to solve it completely. You must solve the complete question all 6 parts. Otherwise you can skip and let others solve it. But don't solve it partially one or three questions. If you do it, I will dislike for sure.
m1: 1 kg m2: 2 kg V1i: 1 m/s V2i: -2 m/s Find the following values from the simulation V1f. -3.00 V2f. O Verify that momentum is conserved and kinetic energy is conserved by completing the following table: Momentum after Kinetic energy Momentum before collision Kinetic energy after collision collision before collision m1V1i + m2V2i m1Vif + m2V2f 2 mivir + ½ m2v21? 2 miV1 + ½ m2V2r
1. If two particles have equal kinetic energies, are their momenta necessarily equal? Explain. 2. If two particles have equal momenta, are their kinetic energies necessarily equal? Explain. 3. An isolated system is initially at rest. Is it possible for parts of the system to be in motion at some later time? If so, explain how this might occur. 4. If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision? Is it possible for one to be at rest after the collision? Explain. 5. Is it possible to have a collision in which all of the kinetic energy is lost? If so, cite an example. 6. Consider a perfectly inelastic collision between a car and a large truck. Which vehicle loses more kinetic energy as a result of the collision? 7. A piece of mud is thrown against a brick wall and sticks to the wall. What happens to the momentum of the mud? Is momentum conserved? Explain.
19
The impulse experienced by a body is proportional to its change in: velocity squared O momentum Okinetic energy O both 1 and 2
How are momentum and impulse related?
Asap plzzzz
Multiple choice, Write the letter of your answer. This is all about General Physics 1: Impulse and Momentum.
Needs Complete typed solution with 100 % accuracy.
Saved Help Required information of 2 Particle A has a mass of 5.40 g and particle B has a mass of 1.80 g. Particle A is located at the origin and particle B is at the point (x, y) = (25.0 cm, 1.70 cm). What is the x-component of the CM? ook cm Hint rint rences 10 11 of 21 rch...pdf 2 Edje PreLab Arch..pdf