An atomic nucleus suddenly bursts apart (fission) into two pieces. Piece A with mass ma travels to the left withPiece B with mass mg travels to the right with speed vB. Show the velocity of piece B in terms ofa speed ofVA.ma, mB and VA.Solution:Consider that the nucleus is not acted by an external force. Thus, momentum is conserved, so:PBf +PAf= 0Substituting the expression for momentum results tomBVmAV= 0Deriving the expression for the velocity of piece B results toV=(mA/mVA

Name: An atomic nucleus suddenly bursts apart (fission) into two pieces. Pi
Uploaded: 2024-03-20T06:57:37.000Z
Channel: Bartleby
Description: An atomic nucleus suddenly bursts apart (fission) into two pieces. Piece A with mass ma travels to the left withPiece B with mass mg travels to the right with speed vB. Show the velocity of piece B in terms ofa speed ofVA.ma, mB and VA.Solution:Cons

The initial speed of an atom will be zero because it is in the rest condition before the fission.…

Science

Physics

An atomic nucleus suddenly bursts apart (fission) into two pieces. Piece A with mass ma travels to the left with a speed of vĄ. Piece B with mass mg travels to the right with speed vg. Show the velocity of piece B in terms of ma mB and VA. Solution: Consider that the nucleus is not acted by an external force. Thus, momentum is conserved, so: PBf + PAf = 0 Substituting the expression for momentum results to mBv + MAV = 0 Deriving the expression for the velocity of piece B results to =(mA/m VA

An atomic nucleus suddenly bursts apart (fission) into two pieces. Piece A with mass ma travels to the left with a speed of vĄ. Piece B with mass mg travels to the right with speed vg. Show the velocity of piece B in terms of ma mB and VA. Solution: Consider that the nucleus is not acted by an external force. Thus, momentum is conserved, so: PBf + PAf = 0 Substituting the expression for momentum results to mBv + MAV = 0 Deriving the expression for the velocity of piece B results to =(mA/m VA

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

A car moving at 12 m/s crashes into a tree and stops in 0.22 s. Randomized Variables v = 12 m/s t = 0.22 s m = 68 kg F= 3709.1 Part (a) What would the final velocity be, in meters per second, if the car with the 68-kg passenger had collided with (and stuck to) a car that has a mass equal to and is traveling in the opposite direction and at the same speed?
A particle of mass m is moving with speed 2v and collides with a mass 2m moving with speed v in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass m, which move at angle 45° with respect to the original direction. The speed of each of the moving particle will be (a) √2 v (d) 2√√2 v (b) V √2 V (Ⓒ) (2√2)
Some particles are created via collisions. For instance, pions are created via collisions with high energy protons and carbon, which releases neutrons and pions (which they themself, decay into muons). Suppose two particles with momenta of pi = (2, -1,0,0) kg and p2 = (4,1, 1,0) kg collide. Three particles are created as a result from this collision, two of which have momenta of P3 = (1,0.5, 0, 0) kg and p4 = (1, -1,0,0) kg, determine the four momentum, energy, rest mass and three velocity of the third daughter particle. %3D
Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time, when they're not sleeping or eating, joyfully scampering about on the cage's floor. Bryce tracks his mice's health diligently and just now recorded their masses as 0.0197 kg, 0.0193 kg, 0.0161 kg, and 0.0151 kg. At this very instant, the x‑ and y‑components of the mice's velocities are, respectively, (0.219 m/s,−0.493 m/s)(, (−0.631 m/s,−0.925 m/s), (0.733 m/s,0.355 m/s), and (−0.489 m/s,0.905 m/s). Calculate the x‑ and y‑components of Bryce's mice's total momentum, px and py.
→ A student runs an experiment with two carts on a low-friction track. As measured in the Earth reference frame, cart 1 (m 0.36 kg) moves from left to right at 1.0 m/s as the student walks along next to it at the same velocity. (a) What velocity v E2,¡ in the Earth reference frame must cart 2 (m 0.12 kg) have before the collision if, in the student's reference frame, cart 2 comes to rest right after the collision and cart 1 reverses direction and travels from right to left at 0.33 m/s? (b) What does the student measure for the momentum of the two-cart system? (c) What does a person standing in the Earth reference frame measure for the momentum of each cart before the collision? .. =
how to do?
My computer uses 15 W, if I run it for 3 hours how many Joules have I used?
Show that the kinetic energy of a particle of mass m is related to the momentum p of that particleby, ? =p^2/2m. (Hint: plug mv in for p and simplify!)
particle of mass m = 5 kg has total energy (rest energy and kinetic energy) of √10mc². What is the momentum of the particle? Give answer in units of kg. m/s
For all the collisions in this lab, what can you conclude about the sum of the changes in momentum for the objects (Δp1 + Δp2 for Parts 1 and 2 and Δp1 + Δp2 + Δp3 for Part 3)? Does this agree with the Law of Momentum Conservation? Show why or why not.
2.2.1) If in the above collision the masses of the particles are m1 = m and m2= 4m and u1/u2 = 3. Show that the opening anglebetween the paths of the m1 and m2 is 120⁰ 2.2.2) If the mass m1 has incident kinetic energy E find the energies of the emerging particles m1 and m2 in terms of E.
The equation a = Fnet / m is valid in any dimension. That is, a = Fnet / m implies ax = Fnet x / m and ay = Fnet y / m. Similarly, the law of conservation of momentum is valid in any dimension. pA0 + pB0 = pA + pB implies pA0x + pB0x = pAx + pBx and pA0y + pB0y = pAy + pBy.Standard cue balls have masses of 170 grams, while numbered billiard balls have masses of only 160 grams. A cue ball traveling at 9.65 meters per second and 0.0 degrees strikes the nine ball at rest. After the collision, the cue ball travels at 3.95 meters per second and 60.7 degrees. What is the velocity of the nine ball after the collision? Include units in your answers.