Answered: Two particles collide, one of them… | bartleby

Related questions

Question

Two particles collide, one of them initially being at rest. Is it possible for both particles to be at
rest after the collision? *
If the collision is perfectly inelastic, then this happens.
O If the collision is elastic, then this happens.
This can happen sometimes if the more massive particle was at rest.
O No.

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 4 steps

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

1. In a perfectly inelastic collision, two objects collide and stick together. Is momentum conserved in the collision? Is energy conserved? Why or why not? A simple example might help.
A projectile has undergone an elastic one-dimensional collision with an initially stationary target, along an x axis. The figure below is a graph of position versus time for the projectile and target, before and after the collision. (Two line segments are parallel to the time axis.) Which is true about the masses of the projectile and target? 4 O the projectile's mass is greater O the target's mass is greater O the masses are equal
Block 1 of mass 200 kg slides over a frictionless surface with a velocity of 0.25 cm/s and strikes block 2 of mass 100 kg sliding to the left at 0.75 cm/s. What is the final velocit y of each block if the collision is A) Perfectly elastic? B) Perfectly inelastic? C) If there is an external force of 4000 kg*cm/s^2 to the right for 0.008 s during the perfectly inelastic
1. A billiard ball, labeled 1, moving horizontally strikes another billiard ball, labled 2, at rest. Before the impact ball 1 was moving at a speed of 5.0000 m/s and after the impact it is moving at a speed of 3.0000 m/s at an angle of 50 degrees from the original direction. The two balls have equal mass. Vii O Ball 1 only O Ball 2 only m₁ y m₂ 0₂ Vif m₁ m/s m₂ V2f (a) What should you consider a system for the collisional forces to be internal in order to apply the law of conservation of momentum ? X O Ball 1 and Ball 2 (b) Write conservation of momentum equations for the x and y directions separately and solve for the speed of ball 2 after the impact. Speed of ball 2 after impact =
Object 1 with a mass of 4.70 kg, moving right at a speed of 7.80 m/s along a frictionless surface, collides head-on with Object 2 with a mass of 3.30 kg moving left at a speed of 10.6 m/s. After the collision, the Object 1 is moving left at 2.40 m/s. 10. What is true about the total momentum before and after the collision? O Equal in magnitude but opposite in direction O Different in magnitude and direction O Equal in magnitude and direction O Different in magnitude but in the same direction
In an elastic collision, one ball has initial velocity v when it collides head-on with another, much less massive ball, initially at rest. The velocity of the second ball after the collision is approximately equal to Multiple Choice v/2 V none of the above 2v
ent G An impulse of 12.2 kg*m/s is deli x 361242/take 10 p Question 1 A warehouse employee is pushing a 30.0 kg desk across a floor at a constant speed of 0.50 m/s. What is the magnitude of the momentum of the desk? O 0.017 kg*m/s O 15.0 kg*m/s O 60 kg*m/s O 30.5 kg*m/s Question 2 10 pts Norton Antivirus: 19.99/YE Your special price is valid to Google Chrome istudy-helper.co The momentum of a 12kg dog racing to greet its owner has a Norton What is the dog's speed? Install O a Cop
After the ball A hits (or collides with) the ball B, they are both observed to move off at 45° to the x axis,ball A above the x axis and ball B below. So ?′A= 45° and ?′B = −45°.Is the collision elastic? Justify your answer
Lucy is cruising through space in her new spaceship. As she coasts along, a tiny spacebug drifts into her path and bounces off the window. Consider several statements concerning this scenario. Evaluate each statement according to the law of momentum conservation and match it to the appropriate category. Must be true according to the law of momentum conservation Must be false according to the law of momentum conservation Answer Bank Not determined by the law of momentum conservation If the spacebug had stuck to the spaceship instead of bouncing off, momentum would not have been conserved for this interaction. The change in the spacebug's momentum during the collision is greater than the change in the spaceship's momentum. The total chanaa in momentum for this interaction is nem The total momentum of the spaceship and spacebug before the collision is equal to their total momentum afterward. 日示全部 X
A radioactive nucleus at rest decays into a second nucleus, an electron, and a neutrino. The electron and neutrino are emitted at right angles and have momenta of Pe=9.67x10-23 kg. m/s and p = 5.99×10-23 kg. m/s. respectively. (Figure 1) Figure Pnuc nuc +y 0 Onuc Py Pe 1 of 1 +X Part A Determine the magnitude of the momentum of the second (recoiling) nucleus. Express your answer to three significant figures and include the appropriate units. Pnuc= Submit Part B Onuc = Submit μA 4 Provide Feedback Value Request Answer Determine the angle between the momentum of the electron and the momentum of the second (rec Express your answer using three significant figures. [VE| ΑΣΦ 1 → C Request Answer Units 1 ? ?
3. This problem refers to the figure below: mmm There are 3 identical bobs of mass m hanging side-by-side. Two are then lifted to a height of h and released. The collisions in this problem are elastic. Answer the following questions: (a) Let's call the initial and final state immediately before and after the collision, respectively. Write down the initial momentum and kinetic energy of the system in terms of m and h. (b) Show that, after the collision, the left and center bobs rise to height h while the right bob becomes stationary. In particular, show that the center bob acts as if it were swinging freely as a lone pendulum.
A marble with mass ?? = 2.5 [g] moves eastward with speed ?? = 1.75 [m/s]. It then collides with another marble of mass ?? = 1.25 [g] and is moving northward with a speed ?? = 2.30 [m/s]. After the collision, marble ? now moves at 50° north of east, while marble ? moves at 20° south of east. 2 A. What is the speed of marble ? after the collision? B. What is the speed of marble ? after the collision? C. Is energy lost or gained in the system

SEE MORE QUESTIONS