Applying conservation of momentum in x-direction as, m₁(-V₁) ₁ + m₂ (va) ₁² = m₂(v₁),+ms (VB), (VA) 3 MB 15 -8 m/s +5 m/s² (VA) x + (VB) x . (1) Here, (VA)x and (VB)x are the velocity after collision in x direction. e = The expression of coefficient of restitution is, VAX - VBx VAX VBì VA1 Substitute given values in above expression, 0.5 VAX-VBx -VBx = = = = 4.8 m/s. On solving equation (1) and (2), ·? VAX (VA) (v₁), + (v₂), (VB) 5 m/s. 7 m/sVB x = (2) m/s
Applying conservation of momentum in x-direction as, m₁(-V₁) ₁ + m₂ (va) ₁² = m₂(v₁),+ms (VB), (VA) 3 MB 15 -8 m/s +5 m/s² (VA) x + (VB) x . (1) Here, (VA)x and (VB)x are the velocity after collision in x direction. e = The expression of coefficient of restitution is, VAX - VBx VAX VBì VA1 Substitute given values in above expression, 0.5 VAX-VBx -VBx = = = = 4.8 m/s. On solving equation (1) and (2), ·? VAX (VA) (v₁), + (v₂), (VB) 5 m/s. 7 m/sVB x = (2) m/s
Classical Dynamics of Particles and Systems
5th Edition
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Stephen T. Thornton, Jerry B. Marion
Chapter9: Dynamics Of A System Of Particles
Section: Chapter Questions
Problem 9.32P
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Question
Can you help me find Vax and Vbx, thank you.
![Applying conservation of momentum in x-direction as,
3
m₁(-v₁)₁ + m₂ (va) ₁² = m₂ (va), +ms (vo),
-VA) MB
(VA)
MA
(VB)
15
-8 m/s+5 m/s²
(VA) x + (VB) x
Here, (VA)x and (VB)x are the velocity after collision in x direction.
e =
The expression of coefficient of restitution is,
VAX-VBx
VAX
VBì VA1
Substitute given values in above expression,
0.5
VAX - VBx
-VBx
=
=
=
= 4.8 m/s.
On solving equation (1) and (2),
·?
VAX
(VA) x + (VB) x
5 m/s.
. (1)
7 m/sVB x
=
(2)
m/s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8eaaea92-e247-4966-90e2-8af37fd7e5c3%2Fd7693946-d548-4d80-b9ac-d5e302bf5a49%2Fvahi2m_processed.png&w=3840&q=75)
Transcribed Image Text:Applying conservation of momentum in x-direction as,
3
m₁(-v₁)₁ + m₂ (va) ₁² = m₂ (va), +ms (vo),
-VA) MB
(VA)
MA
(VB)
15
-8 m/s+5 m/s²
(VA) x + (VB) x
Here, (VA)x and (VB)x are the velocity after collision in x direction.
e =
The expression of coefficient of restitution is,
VAX-VBx
VAX
VBì VA1
Substitute given values in above expression,
0.5
VAX - VBx
-VBx
=
=
=
= 4.8 m/s.
On solving equation (1) and (2),
·?
VAX
(VA) x + (VB) x
5 m/s.
. (1)
7 m/sVB x
=
(2)
m/s
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