Given: • Mass of particle 1 = 0.35 kg, starting velocity of particle 1 = 2.1 m/s Mass of particle 2 = 2.10 kg, starting velocity of particle 2 = -1.1 m/s • coefficient of restitution = 0.40 (elasticity percent) Find: . Velocity of particle 2 after collision =________ m/s • + if moving to the right, - if moving to the left

Given: • Mass of particle 1 = 0.35 kg, starting velocity of particle 1 = 2.1 m/s Mass of particle 2 = 2.10 kg, starting velocity of particle 2 = -1.1 m/s • coefficient of restitution = 0.40 (elasticity percent) Find: . Velocity of particle 2 after collision =________ m/s • + if moving to the right, - if moving to the left

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

Oblique Impact To analyze an oblique impact using the conservation of momentum and coefficient of restitution. When an oblique impact occurs between two smooth particles, the particles move away from each other with velocity vectors that have unknown directions and unknown magnitudes. If the y axis is within the plane of contact and the x axis is the line of impact, the impulsive forces of deformation and restitution act only along the line of impact (the x axis). Momentum of the system is conserved along the line of impact (the x axis): ∑m(vx)1=∑m(vx)2 The coefficient of restitution, e, relates the relative-velocity components of the particles along the line of impact (the x axis): e=(vBx)2−(vAx)2(vAx)1−(vBx)1 The momenta of both particles A and B are conserved in the plane of contact (the y axis) because no impulse acts on either particle in this plane. Therefore, the y component of the velocities before and after the collisions remains unchanged: Immediately after the collision,…
Oblique Impact To analyze an oblique impact using the conservation of momentum and coefficient of restitution. When an oblique impact occurs between two smooth particles, the particles move away from each other with velocity vectors that have unknown directions and unknown magnitudes. If the y axis is within the plane of contact and the x axis is the line of impact, the impulsive forces of deformation and restitution act only along the line of impact (the x axis). Momentum of the system is conserved along the line of impact (the x axis): ∑m(vx)1=∑m(vx)2 The coefficient of restitution, e, relates the relative-velocity components of the particles along the line of impact (the x axis): e=(vBx)2−(vAx)2(vAx)1−(vBx)1 The momenta of both particles A and B are conserved in the plane of contact (the y axis) because no impulse acts on either particle in this plane. Therefore, the y component of the velocities before and after the collisions remains unchanged: (vy)1=(vy)2 As shown,…
A particle (1 kg) moves with 12 m/sec, collide with another particle (2 kg) which moves in the opposite direction by 24 m/sec. If the collision is perfectly elastic, then the movement of each particle after collision are
Students are given two spherical balls made of different materials, but of identical mass and diameter to experiment with. Each ball is dropped from an initial height, h0. The “sad ball” comes to rest without visibly bouncing when it strikes the tabletop. The “happy ball” bounces up to a final height h, where h< h0 A fellow student wants to use one of the balls in a carnival game where cups are knocked over by a thrown ball. Which ball should they use, happy or sad? Why?
A. Estimate the typical maximum horizontal distance that a person can throw a baseball. (Answer: 40m) B. Based on your estimate from part A. What is the speed of the baseball as it leaves the hand? v0= ? (m/s) C. If you could throw the baseball straight up at the same speed as in part B, how high would it go? hmax = ? D. The value of g on Venus is 8.9 m/s2. What horizontal distance can you throw a baseball on Venus if you throw it with the same initial speed as in part B? Rmax=?
n a perfect elastic collision, speed of the approach and velocity of separation are
Restitution coefficient through the measurement of the height of the bounces of a bouncy ball, find the coefficient of restitution between the floor and the ball
A particle (1 kg) moves with 12 m/sec, collide with another particle (2 kg) which moves in the opposite direction by 24 m/sec. When the two particles stick together, then the movement of each particle after collision
A volleyball is hit when it is 4 ft above the groundand 12 ft from a 6-ft-high net. It leaves the point of impact withan initial velocity of 35 ft / sec at an angle of 27 and slips by theopposing team untouched. Suppose that the net is raised to 8 ft. Does this change things? Explain.
Two rods, with masses MA and MB having a coefficient of restitution, e, move along a common line on a surface, figure 2. a) Find the general expression for the velocities of the two rods after impact. b) IfmA =2kg,mB =1kg,vB=3m/s,ande=0.65,find the value of the initial velocity vA of rod A for it to be at rest after the impact and the final velocity v’B of rod B. c) Find the percent decrease in kinetic energy which corresponds to the impact in part b
A 35 g steel ball is held by a ceiling-mounted electromagnet 3.4 m above the floor. A compressed-air cannon sits on the floor, 5.0 m to one side of the point directly under the ball. When a button is pressed, the ball drops and, simultaneously, the cannon fires a 25 g plastic ball. The two balls collide 1.2 m above the floor. What was the launch speed of the plastic ball?
Required minimum velocity (m/s) of the particle maximum velocity (m/s) if maximum tension does not exceed 141.12 N