Problem 2-2D collisions. Two particles of equal mass collide. Before the collision, particle A has initial velocity VA, while particle B is at rest. After the collision, the angle of deflection of particle A is 04-30°. Assume, at first (in parts a and b), that the collision is elastic. (a) Find the angle of deflection of particle B. (b) Assume that the velocity of particle A before the collision is VA= 100 m/s and that the kinetic energy of particle B after the collision is Ks' = 20 J. Find the particles' mass. (c) If, instead, the collision is fully inelastic, what is the angle of deflection and the velocity of the combined A+B particle after the collision?
Problem 2-2D collisions. Two particles of equal mass collide. Before the collision, particle A has initial velocity VA, while particle B is at rest. After the collision, the angle of deflection of particle A is 04-30°. Assume, at first (in parts a and b), that the collision is elastic. (a) Find the angle of deflection of particle B. (b) Assume that the velocity of particle A before the collision is VA= 100 m/s and that the kinetic energy of particle B after the collision is Ks' = 20 J. Find the particles' mass. (c) If, instead, the collision is fully inelastic, what is the angle of deflection and the velocity of the combined A+B particle after the collision?
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Question
Can you explain part a and b of this question
Specifically why in part A is it -60 and not +60
And part b how does the sin become cos
And when i substitute the values I never get radical 3 please show me
![Vp + Voy
Problem 2-2D collisions.
Two particles of equal mass collide. Before the collision, particle A has initial velocity VA, while particle B is at rest. After the collision, the angle of deflection
of particle A is 04-30°. Assume, at first (in parts a and b), that the collision is elastic. (a) Find the angle of deflection of particle B. (b) Assume that the velocity
of particle A before the collision is VA= 100 m/s and that the kinetic energy of particle B after the collision is KB' = 20 J. Find the particles' mass. (c) If, instead,
the collision is fully inelastic, what is the angle of deflection and the velocity of the combined A+B particle after the collision?
articles of canal
#1
mass
elideen](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2F43c1b00e-4cb5-4573-bf3a-a52384473afe%2Fctwjeen_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Vp + Voy
Problem 2-2D collisions.
Two particles of equal mass collide. Before the collision, particle A has initial velocity VA, while particle B is at rest. After the collision, the angle of deflection
of particle A is 04-30°. Assume, at first (in parts a and b), that the collision is elastic. (a) Find the angle of deflection of particle B. (b) Assume that the velocity
of particle A before the collision is VA= 100 m/s and that the kinetic energy of particle B after the collision is KB' = 20 J. Find the particles' mass. (c) If, instead,
the collision is fully inelastic, what is the angle of deflection and the velocity of the combined A+B particle after the collision?
articles of canal
#1
mass
elideen
![sics | - Written exam 14-07-2021 temp.docx
Problem 2 - 2D collisions.
(a) For a 2D elastic collisions between two
particles of equal mass, the angle between
and is ± 11/2. Therefore, it 0₁ = 30°
then = -60°
A before the collusion
(b) Assuming that particle
y-component
moves along the x-axis, the
the equation for linear momentum conservation
of
15
10 = V sin 0₁
+
Therefore:
Y sin B
83
V₁ = - in V₂ = √3 V6
COS PA
Conservation of kinetic energy gives
12
√ ² = √₁² + √₂ ² = 4√₂²"
12
2
A VB
4.VB
V = VA
4
2
2
MY V K.
B
and so
K
тв
=
8kB
12
A
=
8x205
42
1x 10 m/s²
-2
1.6×10 * 12
MA=MB = 16 g
(C) If the collision is totally inelastic, then
the combined object A+B, with a total
mass MA + m₂ = 2mA, must move along the
2-direction. Conservation of linear momentum
gives:
MAVA - MA+M₂ →
V = VA= 50m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2F43c1b00e-4cb5-4573-bf3a-a52384473afe%2Fl0tye4l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:sics | - Written exam 14-07-2021 temp.docx
Problem 2 - 2D collisions.
(a) For a 2D elastic collisions between two
particles of equal mass, the angle between
and is ± 11/2. Therefore, it 0₁ = 30°
then = -60°
A before the collusion
(b) Assuming that particle
y-component
moves along the x-axis, the
the equation for linear momentum conservation
of
15
10 = V sin 0₁
+
Therefore:
Y sin B
83
V₁ = - in V₂ = √3 V6
COS PA
Conservation of kinetic energy gives
12
√ ² = √₁² + √₂ ² = 4√₂²"
12
2
A VB
4.VB
V = VA
4
2
2
MY V K.
B
and so
K
тв
=
8kB
12
A
=
8x205
42
1x 10 m/s²
-2
1.6×10 * 12
MA=MB = 16 g
(C) If the collision is totally inelastic, then
the combined object A+B, with a total
mass MA + m₂ = 2mA, must move along the
2-direction. Conservation of linear momentum
gives:
MAVA - MA+M₂ →
V = VA= 50m
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