Problem 2-2D Collisions (a) A particles of mass m₁ = 5 kg collides with another particle of mass mg. The collision is observed in the centre-of-mass (CM) reference frame. In this frame, particle A has a kinetic energy 1 MJ and the particle B has 5 MJ. Find the velocity of particle B in the CM frame before collision. (b) Assume that the collision is totally inelastic. Find the difference in the total kinetic energy, Kafter-Kbefore, and the momentum of the combined particle A+B after the collision, as observed in the CM frame. (c) Finally, assume that the collision is elastic, and the trajectory of particle B after the collision, as observed in the CM frame, forms an angle 0B-30° with the original axis of motion of the two particles before collision; find the angle OA and the speed VA of particle A after the collision. = Y = 24 g = 24 TD each other, So they have velocity -2-vectors anti-parallel. Test-2 MV = 2×5×106 = 107 = Let us consider before collision, V = VA 2 B =- VB 2 Then, total momentum, MA VA+(V) 2 = 0; MAVA = mBVB It is given that, MAVA² = 1 MJ, MA= 5 kg and ½ MAVZ == Imavg = ㅎ 2 MAVA b) The total kinetic energy collision Kbefore = ½ 107 :. VB = = 10 = 10% m/s before =(1+5) MJ = 6 MJ Since it is an inelastic collision, the total k. I. after collison is zero in the CM frame. , mv = 5 mV = 5x2x10³] So, OK = Kaffer before = - 6 MJ as (MAVA). VS = 10* (using mave my From ½ m² = 1×10 2 m² V₁² = 2X 106 M²AVA ΜΑ c) 9A B
Problem 2-2D Collisions (a) A particles of mass m₁ = 5 kg collides with another particle of mass mg. The collision is observed in the centre-of-mass (CM) reference frame. In this frame, particle A has a kinetic energy 1 MJ and the particle B has 5 MJ. Find the velocity of particle B in the CM frame before collision. (b) Assume that the collision is totally inelastic. Find the difference in the total kinetic energy, Kafter-Kbefore, and the momentum of the combined particle A+B after the collision, as observed in the CM frame. (c) Finally, assume that the collision is elastic, and the trajectory of particle B after the collision, as observed in the CM frame, forms an angle 0B-30° with the original axis of motion of the two particles before collision; find the angle OA and the speed VA of particle A after the collision. = Y = 24 g = 24 TD each other, So they have velocity -2-vectors anti-parallel. Test-2 MV = 2×5×106 = 107 = Let us consider before collision, V = VA 2 B =- VB 2 Then, total momentum, MA VA+(V) 2 = 0; MAVA = mBVB It is given that, MAVA² = 1 MJ, MA= 5 kg and ½ MAVZ == Imavg = ㅎ 2 MAVA b) The total kinetic energy collision Kbefore = ½ 107 :. VB = = 10 = 10% m/s before =(1+5) MJ = 6 MJ Since it is an inelastic collision, the total k. I. after collison is zero in the CM frame. , mv = 5 mV = 5x2x10³] So, OK = Kaffer before = - 6 MJ as (MAVA). VS = 10* (using mave my From ½ m² = 1×10 2 m² V₁² = 2X 106 M²AVA ΜΑ c) 9A B
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)...
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I attached both the question and answers
Usually for center of mass reference frame what I do is mAvA + mBvB =0
Because the velocity of B should be negative I.e counter parallel to that of A
Transcribed Image Text:Problem 2-2D Collisions
(a) A particles of mass m₁ = 5 kg collides with another particle of mass mg. The collision is observed in the centre-of-mass (CM) reference frame. In this
frame, particle A has a kinetic energy 1 MJ and the particle B has 5 MJ. Find the velocity of particle B in the CM frame before collision. (b) Assume that the
collision is totally inelastic. Find the difference in the total kinetic energy, Kafter-Kbefore, and the momentum of the combined particle A+B after the collision, as
observed in the CM frame. (c) Finally, assume that the collision is elastic, and the trajectory of particle B after the collision, as observed in the CM frame,
forms an angle 0B-30° with the original axis of motion of the two particles before collision; find the angle OA and the speed VA of particle A after the collision.
=
![Y = 24 g = 24 TD
each other, So they have velocity
-2-vectors anti-parallel.
Test-2
MV = 2×5×106 = 107
=
Let us consider before collision,
V = VA 2 B =- VB 2
Then, total momentum,
MA VA+(V) 2 = 0;
MAVA = mBVB
It is given that,
MAVA² = 1 MJ, MA= 5 kg
and ½ MAVZ
==
Imavg = ㅎ
2
MAVA
b) The total kinetic energy
collision Kbefore = ½
107
:. VB
=
=
10 = 10% m/s
before
=(1+5) MJ
= 6 MJ
Since it is an inelastic collision,
the total k. I. after collison is
zero in the CM frame.
, mv = 5 mV = 5x2x10³] So, OK = Kaffer before = - 6 MJ
as (MAVA). VS = 10* (using mave my
From ½ m² = 1×10
2
m² V₁² = 2X 106
M²AVA
ΜΑ
c)
9A
B](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2F98fd038f-8a3b-4fed-bb4b-2c9afe29d53d%2Fi14ifmt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Y = 24 g = 24 TD
each other, So they have velocity
-2-vectors anti-parallel.
Test-2
MV = 2×5×106 = 107
=
Let us consider before collision,
V = VA 2 B =- VB 2
Then, total momentum,
MA VA+(V) 2 = 0;
MAVA = mBVB
It is given that,
MAVA² = 1 MJ, MA= 5 kg
and ½ MAVZ
==
Imavg = ㅎ
2
MAVA
b) The total kinetic energy
collision Kbefore = ½
107
:. VB
=
=
10 = 10% m/s
before
=(1+5) MJ
= 6 MJ
Since it is an inelastic collision,
the total k. I. after collison is
zero in the CM frame.
, mv = 5 mV = 5x2x10³] So, OK = Kaffer before = - 6 MJ
as (MAVA). VS = 10* (using mave my
From ½ m² = 1×10
2
m² V₁² = 2X 106
M²AVA
ΜΑ
c)
9A
B
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