10/20/2017 Homework: Collisions, Impulse, and Reference Frames Friday, October 2
ID: 1783993 • Letter: 1
Question
10/20/2017 Homework: Collisions, Impulse, and Reference Frames Friday, October 208:28 AM athars@dupage.eduaccount log off hep Flpthysics hysis211 Fall 2017 Physics 2111 Fall 2017 College of Dupage Unit 13: Prelecture CheckpointHomework/ Problems Homework: Collisions, Impulse, And Reference Frames Print Assignment View Deadline: 100% until Friday, October 20 at 9:30 AM Ball Hits wai (w/ Work) Explosion (w/ Work) 12 3 456 7 Everose Optional andard Eaerose all Hits Wall 2 Explosion (w/ Work) Optional Randard Exercse pN Scattering Randerd Exercse Wooden Blocks in the Center of Mass (COM) frame An object with total mass metal-14.9 kg is sitting at rest when it explodes into three pieces. One piece with mass m 4.8 kg moves up and to the left at an angle of 1-23. above the-X axis with a speed of v1-28 m/s. A second piece with mass m2-5.2 kg moves down and to the right an angle of 2-28. to the right of the-y axis at a speed of v2-21 m/s 1) What is the magnitude of the final momentum of the system (all three pieces? kg-m/s Submit Help 2) What is the mass of the third piece? kg Submit Help 3) what is the x-component of the velacity of the third piece? m/s Submit Help 4) What is the y-component of the velocity of the third piece? m/s Submit 5) What is the magnitude of the velocity of the center of mass of the pieces after the collision? m/s Submit 6) Calculate the increase in kinetic energy of the pieces during the explosion. https://www.flipitphysics.com/Course/ViewProblem?unititemlD-30582558enrollmentlD-254559 1/2Explanation / Answer
m = 14.9 kg
m1 = 4.8 kg
m2 = 5.2 kg
m3 = 14.9 - 4.8 - 5.2 = 4.9 kg
v1 = 28(-cos23 i^ + sin23 j^)
v2 = 21(sin28 i^ - cos28 j^)
as initially m is at rest, net momentu is zero which will be zero at the end as well.
m1v1 + m2v2 + m3v3 = 0
4.8 x 28(-cos23 i^ + sin23 j^) + 5.2 x 21(sin28 i^ - cos28 j^) + 4.9v3 = 0
=> v3 = (4.8 x 28(cos23 i^ - sin23 j^) + 5.2 x 21(-sin28 i^ + cos28 j^))/4.9
= 14.7856 i^ + 8.96 j^
|v3| = 17.29 m/s
1) net momemtum of system = 0
2) mass of third mass = 4.9 kg
3) x component of velocity of 3rd piece = 14.7856 m/s
4) y component of velocity of 3rd piece = 8.96 m/s
5) velocity of center of mass = 0
6) kinetic energy increase = 0.5(m1v12 + m2v22 + m3v32) = 0.5(4.8 x 28 x 28 + 5.2 x 21 x 21 + 4.9 x 17.29 x 17.29)
= 3760.61 J
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