Applying conservation of momentum in x-direction as, MA mA(vA), +ms(vs) (VA)₂ + (VB) x 5 m/s...... (1) 3 MA ( - VA) ₁ + MB (VB) ₁³/ А 1 15 -8 m/s+5m/s²/ (VA) x + (VB)x Here, (VA)X and (VB)x are the velocity after collision in x direction. The expression of coefficient of restitution is, VAX-VBx VB₁-VA1 e Substitute given values in above expression, VAX - VBx 0.5 5--5 5 m/s......... (2) VAX VBx || On solving equation (1) and (2), VAX 7 m/SVBx ? m/s
Applying conservation of momentum in x-direction as, MA mA(vA), +ms(vs) (VA)₂ + (VB) x 5 m/s...... (1) 3 MA ( - VA) ₁ + MB (VB) ₁³/ А 1 15 -8 m/s+5m/s²/ (VA) x + (VB)x Here, (VA)X and (VB)x are the velocity after collision in x direction. The expression of coefficient of restitution is, VAX-VBx VB₁-VA1 e Substitute given values in above expression, VAX - VBx 0.5 5--5 5 m/s......... (2) VAX VBx || On solving equation (1) and (2), VAX 7 m/SVBx ? m/s
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Can you help me find VAx and VBx
![## Conservation of Momentum in the X-Direction
### Applying Conservation of Momentum
Applying conservation of momentum in the x-direction for two colliding bodies, we have:
\[
m_A (-v_{A}) + m_B (v_B) \frac{3}{15} = m_A (v_A)_x + m_B (v_B)_x
\]
This simplifies to:
\[
-8 \, \text{m/s} + 5 \, \text{m/s} \cdot \frac{3}{5} = (v_A)_x + (v_B)_x = -5 \, \text{m/s} \quad \text{(1)}
\]
Here, \((v_A)_x\) and \((v_B)_x\) are the velocities after the collision in the x direction.
### Coefficient of Restitution
The expression for the coefficient of restitution \(e\) is given by:
\[
e = \frac{v_{Ax} - v_{Bx}}{v_{B1} - v_{A1}}
\]
### Substituting Values
Substitute the given values into the expression for restitution:
\[
0.5 = \frac{v_{Ax} - v_{Bx}}{5 - 5}
\]
This leads to:
\[
v_{Ax} - v_{Bx} = 5 \, \text{m/s} \quad \text{(2)}
\]
### Solving the Equations
On solving equations (1) and (2):
\[
v_{Ax} = \, ? \, \text{m/s} \quad v_{Bx} = \, ? \, \text{m/s}
\]
Note: The question marks indicate that specific numerical values are not provided in the image and need to be calculated or given.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8eaaea92-e247-4966-90e2-8af37fd7e5c3%2F6292e6d9-4f2f-4027-8f84-b87b576b027c%2Fxz94y2f_processed.png&w=3840&q=75)
Transcribed Image Text:## Conservation of Momentum in the X-Direction
### Applying Conservation of Momentum
Applying conservation of momentum in the x-direction for two colliding bodies, we have:
\[
m_A (-v_{A}) + m_B (v_B) \frac{3}{15} = m_A (v_A)_x + m_B (v_B)_x
\]
This simplifies to:
\[
-8 \, \text{m/s} + 5 \, \text{m/s} \cdot \frac{3}{5} = (v_A)_x + (v_B)_x = -5 \, \text{m/s} \quad \text{(1)}
\]
Here, \((v_A)_x\) and \((v_B)_x\) are the velocities after the collision in the x direction.
### Coefficient of Restitution
The expression for the coefficient of restitution \(e\) is given by:
\[
e = \frac{v_{Ax} - v_{Bx}}{v_{B1} - v_{A1}}
\]
### Substituting Values
Substitute the given values into the expression for restitution:
\[
0.5 = \frac{v_{Ax} - v_{Bx}}{5 - 5}
\]
This leads to:
\[
v_{Ax} - v_{Bx} = 5 \, \text{m/s} \quad \text{(2)}
\]
### Solving the Equations
On solving equations (1) and (2):
\[
v_{Ax} = \, ? \, \text{m/s} \quad v_{Bx} = \, ? \, \text{m/s}
\]
Note: The question marks indicate that specific numerical values are not provided in the image and need to be calculated or given.
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