Question
A ball of mass 10kg moving with a speed of 8ms^-1collides with another ball mass 20kg initially at rest. After the collision A and B move in directions making angles of 30° and 45° respectively with the initial direction of motion of A. Calculate the speeds of A and B after the collision
Answers
GPT-4o mini
To solve this problem, we can use the principle of conservation of linear momentum. Given:
- Mass of ball A (\(m_A\)) = 10 kg
- Initial speed of ball A (\(v_{A_i}\)) = 8 m/s
- Mass of ball B (\(m_B\)) = 20 kg
- Initial speed of ball B (\(v_{B_i}\)) = 0 m/s (initially at rest)
After the collision:
- Let the speed of ball A after the collision be (\(v_{A_f}\)).
- Let the speed of ball B after the collision be (\(v_{B_f}\)).
- Angle of ball A after collision = 30° from the initial direction of motion.
- Angle of ball B after collision = 45° from the initial direction of motion.
### Step 1: Conservation of Momentum in the x-direction
The total momentum before the collision in the x-direction is:
\[
p_{x_i} = m_A v_{A_i} + m_B v_{B_i} = 10 \times 8 + 20 \times 0 = 80 \, \text{kg m/s}
\]
The total momentum after the collision in the x-direction is:
\[
p_{x_f} = m_A v_{A_f} \cos(30^\circ) + m_B v_{B_f} \cos(45^\circ)
\]
Using conservation of momentum in the x-direction:
\[
80 = 10 v_{A_f} \cos(30^\circ) + 20 v_{B_f} \cos(45^\circ)
\]
### Step 2: Conservation of Momentum in the y-direction
The total momentum before the collision in the y-direction is:
\[
p_{y_i} = 0 \, \text{kg m/s} \quad \text{(as both balls were moving horizontally)}
\]
The total momentum after the collision in the y-direction is:
\[
p_{y_f} = m_A v_{A_f} \sin(30^\circ) - m_B v_{B_f} \sin(45^\circ)
\]
Using conservation of momentum in the y-direction:
\[
0 = 10 v_{A_f} \sin(30^\circ) - 20 v_{B_f} \sin(45^\circ)
\]
### Step 3: Simplifying the equations
1. From the x-direction equation:
\[
80 = 10 v_{A_f} \cdot \frac{\sqrt{3}}{2} + 20 v_{B_f} \cdot \frac{\sqrt{2}}{2}
\]
This simplifies to:
\[
80 = 5\sqrt{3} v_{A_f} + 10\sqrt{2} v_{B_f} \quad (1)
\]
2. From the y-direction equation:
\[
10 v_{A_f} \cdot \frac{1}{2} = 20 v_{B_f} \cdot \frac{\sqrt{2}}{2}
\]
This simplifies to:
\[
5 v_{A_f} = 10\sqrt{2} v_{B_f} \quad (2)
\]
### Step 4: Solve the equations
From equation (2):
\[
v_{A_f} = 2\sqrt{2} v_{B_f}
\]
Substituting \(v_{A_f}\) in equation (1):
\[
80 = 5\sqrt{3} (2\sqrt{2} v_{B_f}) + 10\sqrt{2} v_{B_f}
\]
This simplifies to:
\[
80 = 10\sqrt{6} v_{B_f} + 10\sqrt{2} v_{B_f}
\]
\[
80 = 10 v_{B_f} (\sqrt{6} + \sqrt{2})
\]
Dividing both sides by 10:
\[
8 = v_{B_f} (\sqrt{6} + \sqrt{2})
\]
Thus,
\[
v_{B_f} = \frac{8}{\sqrt{6}+\sqrt{2}}
\]
Now, rationalizing the denominator:
\[
v_{B_f} = \frac{8(\sqrt{6}-\sqrt{2})}{(\sqrt{6})^2 - (\sqrt{2})^2} = \frac{8(\sqrt{6}-\sqrt{2})}{6 - 2} = \frac{8(\sqrt{6}-\sqrt{2})}{4} = 2(\sqrt{6}-\sqrt{2})
\]
Now substitute \(v_{B_f}\) back into \(v_{A_f}\):
\[
v_{A_f} = 2\sqrt{2} v_{B_f} = 2\sqrt{2} \cdot 2(\sqrt{6} - \sqrt{2}) = 4\sqrt{2}(\sqrt{6} - \sqrt{2}) = 4(\sqrt{12} - 2) = 4(2\sqrt{3} - 2)
\]
Thus:
\[
v_{A_f} = 8\sqrt{3} - 8
\]
### Final Answers
So, we have:
- \(v_{A_f} = 4\sqrt{2}(\sqrt{6} - \sqrt{2}) \) m/s
- \(v_{B_f} = 2(\sqrt{6} - \sqrt{2}) \) m/s
Now you can calculate approximate values using numeric values for \(\sqrt{2}\) and \(\sqrt{3}\).
- \(\sqrt{2} \approx 1.414\)
- \(\sqrt{6} \approx 2.449\)
Now plug those into the feasibility of physics and compute the approximate velocities.
- Mass of ball A (\(m_A\)) = 10 kg
- Initial speed of ball A (\(v_{A_i}\)) = 8 m/s
- Mass of ball B (\(m_B\)) = 20 kg
- Initial speed of ball B (\(v_{B_i}\)) = 0 m/s (initially at rest)
After the collision:
- Let the speed of ball A after the collision be (\(v_{A_f}\)).
- Let the speed of ball B after the collision be (\(v_{B_f}\)).
- Angle of ball A after collision = 30° from the initial direction of motion.
- Angle of ball B after collision = 45° from the initial direction of motion.
### Step 1: Conservation of Momentum in the x-direction
The total momentum before the collision in the x-direction is:
\[
p_{x_i} = m_A v_{A_i} + m_B v_{B_i} = 10 \times 8 + 20 \times 0 = 80 \, \text{kg m/s}
\]
The total momentum after the collision in the x-direction is:
\[
p_{x_f} = m_A v_{A_f} \cos(30^\circ) + m_B v_{B_f} \cos(45^\circ)
\]
Using conservation of momentum in the x-direction:
\[
80 = 10 v_{A_f} \cos(30^\circ) + 20 v_{B_f} \cos(45^\circ)
\]
### Step 2: Conservation of Momentum in the y-direction
The total momentum before the collision in the y-direction is:
\[
p_{y_i} = 0 \, \text{kg m/s} \quad \text{(as both balls were moving horizontally)}
\]
The total momentum after the collision in the y-direction is:
\[
p_{y_f} = m_A v_{A_f} \sin(30^\circ) - m_B v_{B_f} \sin(45^\circ)
\]
Using conservation of momentum in the y-direction:
\[
0 = 10 v_{A_f} \sin(30^\circ) - 20 v_{B_f} \sin(45^\circ)
\]
### Step 3: Simplifying the equations
1. From the x-direction equation:
\[
80 = 10 v_{A_f} \cdot \frac{\sqrt{3}}{2} + 20 v_{B_f} \cdot \frac{\sqrt{2}}{2}
\]
This simplifies to:
\[
80 = 5\sqrt{3} v_{A_f} + 10\sqrt{2} v_{B_f} \quad (1)
\]
2. From the y-direction equation:
\[
10 v_{A_f} \cdot \frac{1}{2} = 20 v_{B_f} \cdot \frac{\sqrt{2}}{2}
\]
This simplifies to:
\[
5 v_{A_f} = 10\sqrt{2} v_{B_f} \quad (2)
\]
### Step 4: Solve the equations
From equation (2):
\[
v_{A_f} = 2\sqrt{2} v_{B_f}
\]
Substituting \(v_{A_f}\) in equation (1):
\[
80 = 5\sqrt{3} (2\sqrt{2} v_{B_f}) + 10\sqrt{2} v_{B_f}
\]
This simplifies to:
\[
80 = 10\sqrt{6} v_{B_f} + 10\sqrt{2} v_{B_f}
\]
\[
80 = 10 v_{B_f} (\sqrt{6} + \sqrt{2})
\]
Dividing both sides by 10:
\[
8 = v_{B_f} (\sqrt{6} + \sqrt{2})
\]
Thus,
\[
v_{B_f} = \frac{8}{\sqrt{6}+\sqrt{2}}
\]
Now, rationalizing the denominator:
\[
v_{B_f} = \frac{8(\sqrt{6}-\sqrt{2})}{(\sqrt{6})^2 - (\sqrt{2})^2} = \frac{8(\sqrt{6}-\sqrt{2})}{6 - 2} = \frac{8(\sqrt{6}-\sqrt{2})}{4} = 2(\sqrt{6}-\sqrt{2})
\]
Now substitute \(v_{B_f}\) back into \(v_{A_f}\):
\[
v_{A_f} = 2\sqrt{2} v_{B_f} = 2\sqrt{2} \cdot 2(\sqrt{6} - \sqrt{2}) = 4\sqrt{2}(\sqrt{6} - \sqrt{2}) = 4(\sqrt{12} - 2) = 4(2\sqrt{3} - 2)
\]
Thus:
\[
v_{A_f} = 8\sqrt{3} - 8
\]
### Final Answers
So, we have:
- \(v_{A_f} = 4\sqrt{2}(\sqrt{6} - \sqrt{2}) \) m/s
- \(v_{B_f} = 2(\sqrt{6} - \sqrt{2}) \) m/s
Now you can calculate approximate values using numeric values for \(\sqrt{2}\) and \(\sqrt{3}\).
- \(\sqrt{2} \approx 1.414\)
- \(\sqrt{6} \approx 2.449\)
Now plug those into the feasibility of physics and compute the approximate velocities.