10/20/2017 Homework: Collisions, Impulse, and Reference Frames Friday, October 2

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10/20/2017 Homework: Collisions, Impulse, and Reference Frames Friday, October 208:24 A athars@dupage.eduaccount log off hep FlipltPhysics Physics 2111 Fall 2017 College of Dupage Unit 13: PrelectureCheckpointHomework/ Problems Homework: Collisions, Impulse, And Reference Frames Print Assignment View Dead ne: 100% until Friday, October 20 at 9:30 AM Everose Ball Hits Wall (wl Work) Wooden Blocks in the Center of Mass (COM) frame 12 3 456 7 Everose Optional anderd Exerose all Hits Wall 2 A wooden block with m 2.8kg is sliding across a frictionless surface in the-x direction at 5.3 m/s. A smaller wooden block with m2 -0.9 kg block is traveting in the x direction at 3.2 m/s. Explosion (wl Work) 1) (a) Find the velocity vcm of the center of mass. tanderd terose Explosion 2 m/s Submit Optional Randard Exercse pN Scattering 2) (b) What is the velocity of m in the COM frame? mis Submit Randerd Exercse Wooden Blocks in the Center of Mass (COM) frame 3) What is the velocity of m2 in the COM frame? m/s Submit 4) Now they make a head-on clastic collision. (This means that in the COM frame, the velocity of each is reversed.) What is the velocity of mi in the COM frame after the collision m/s Subnit 5) What is the velocity of m in the COM frame after the collision? mis Submit 6) Transform back into the original frame by adding vn to the velacity of each block. What is the velocity of the mi block in the lab frame after the collision? mis Submit 7) What is the velocity of the m2 block in the lab frame after the collision? m/s Submit Copyright © 2017 Freeman worth Publishers-a dvision of Macmillan Learning. About I Tech Support1 Find Your Local Sales Rep I Terms Of Use I Privacy Policy https://www.flipitphysics.com/Course/ViewProblem?unititemlD-3058257&enrollmentiD-254559;

Explanation / Answer

Given

m1 = 2.8 kg, moving along -x direction with v1 = -5.3 m/s

m2 = 0.9 kg, moving along +x direction with v2 = 3.2 m/s

1.a) velocity of Vcm of the center of mass is  

Vcm = m1v1+m2v2/(m1+m2)

Vcm= (2.8*(-5.3)+0.9*3.2)/(2.8+0.9) m/s

Vcm = -3.23 m/s

2) velocity of m1 in the COM frame is  

V1_cm = V1 - Vcm = -5.3-(-3.23) = -2.07 m/s

3)

velocity of m2 in the COM frame is  

V2_cm = V2 - Vcm = 3.2-(-3.23) = 6.43 m/s

4) head on collision their velocities are reversed then

Vcm = m1v1+m2v2/(m1+m2)

Vcm = (2.8*(5.3)+0.9*(-3.2))/(2.8+0.9) m/s

Vcm = 3.23 m/s

velocity of m1 in the COM frame is  

V1_cm = V1 - Vcm = 5.3-(3.23) = 2.07 m/s

5) velocity of m2 in the COM frame is  

V2_cm = V2 - Vcm = -3.2-(3.23) = -6.43 m/s

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