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To determine whether the collision is elastic, we need to analyze both the kinetic energy and the momentum of the system before and after the collision.
### Step-by-Step Solution:
1. Calculate initial kinetic energy:
First, we need to find the kinetic energy of each mass before the collision.
The formula for kinetic energy ([tex]\( KE \)[/tex]) is:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
where [tex]\( m \)[/tex] is the mass and [tex]\( v \)[/tex] is the velocity.
- For the 3 kg mass:
[tex]\[ KE_{1\_initial} = \frac{1}{2} \times 3 \, \text{kg} \times (4 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{1\_initial} = \frac{1}{2} \times 3 \times 16 \][/tex]
[tex]\[ KE_{1\_initial} = \frac{1}{2} \times 48 \][/tex]
[tex]\[ KE_{1\_initial} = 24 \, \text{J} \][/tex]
- For the 1 kg mass (initially stationary):
[tex]\[ KE_{2\_initial} = \frac{1}{2} \times 1 \, \text{kg} \times (0 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{2\_initial} = 0 \, \text{J} \][/tex]
Total initial kinetic energy:
[tex]\[ KE_{initial} = KE_{1\_initial} + KE_{2\_initial} \][/tex]
[tex]\[ KE_{initial} = 24 \, \text{J} + 0 \, \text{J} \][/tex]
[tex]\[ KE_{initial} = 24 \, \text{J} \][/tex]
2. Calculate final kinetic energy:
Now we find the kinetic energy of each mass after the collision.
- For the 3 kg mass (now moving at 2 m/s):
[tex]\[ KE_{1\_final} = \frac{1}{2} \times 3 \, \text{kg} \times (2 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{1\_final} = \frac{1}{2} \times 3 \times 4 \][/tex]
[tex]\[ KE_{1\_final} = \frac{1}{2} \times 12 \][/tex]
[tex]\[ KE_{1\_final} = 6 \, \text{J} \][/tex]
- For the 1 kg mass (now moving at 6 m/s):
[tex]\[ KE_{2\_final} = \frac{1}{2} \times 1 \, \text{kg} \times (6 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{2\_final} = \frac{1}{2} \times 1 \times 36 \][/tex]
[tex]\[ KE_{2\_final} = \frac{1}{2} \times 36 \][/tex]
[tex]\[ KE_{2\_final} = 18 \, \text{J} \][/tex]
Total final kinetic energy:
[tex]\[ KE_{final} = KE_{1\_final} + KE_{2\_final} \][/tex]
[tex]\[ KE_{final} = 6 \, \text{J} + 18 \, \text{J} \][/tex]
[tex]\[ KE_{final} = 24 \, \text{J} \][/tex]
3. Check if kinetic energy is conserved:
[tex]\[ KE_{initial} = KE_{final} \][/tex]
[tex]\[ 24 \, \text{J} = 24 \, \text{J} \][/tex]
Since the initial and final kinetic energies are equal, the kinetic energy is conserved.
4. Calculate initial and final momentum:
The formula for momentum ([tex]\( p \)[/tex]) is:
[tex]\[ p = mv \][/tex]
where [tex]\( m \)[/tex] is the mass and [tex]\( v \)[/tex] is the velocity.
- Initial momentum:
[tex]\[ p_{initial} = (3 \, \text{kg} \times 4 \, \text{m/s}) + (1 \, \text{kg} \times 0 \, \text{m/s}) \][/tex]
[tex]\[ p_{initial} = 12 \, \text{kg} \cdot \text{m/s} \][/tex]
- Final momentum:
[tex]\[ p_{final} = (3 \, \text{kg} \times 2 \, \text{m/s}) + (1 \, \text{kg} \times 6 \, \text{m/s}) \][/tex]
[tex]\[ p_{final} = (3 \times 2) + (1 \times 6) \][/tex]
[tex]\[ p_{final} = 6 + 6 \][/tex]
[tex]\[ p_{final} = 12 \, \text{kg} \cdot \text{m/s} \][/tex]
5. Check if momentum is conserved:
[tex]\[ p_{initial} = p_{final} \][/tex]
[tex]\[ 12 \, \text{kg} \cdot \text{m/s} = 12 \, \text{kg} \cdot \text{m/s} \][/tex]
Since the initial and final momenta are equal, momentum is conserved.
### Conclusion:
Both kinetic energy and momentum are conserved in this collision. Therefore, the collision is elastic.
### Step-by-Step Solution:
1. Calculate initial kinetic energy:
First, we need to find the kinetic energy of each mass before the collision.
The formula for kinetic energy ([tex]\( KE \)[/tex]) is:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
where [tex]\( m \)[/tex] is the mass and [tex]\( v \)[/tex] is the velocity.
- For the 3 kg mass:
[tex]\[ KE_{1\_initial} = \frac{1}{2} \times 3 \, \text{kg} \times (4 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{1\_initial} = \frac{1}{2} \times 3 \times 16 \][/tex]
[tex]\[ KE_{1\_initial} = \frac{1}{2} \times 48 \][/tex]
[tex]\[ KE_{1\_initial} = 24 \, \text{J} \][/tex]
- For the 1 kg mass (initially stationary):
[tex]\[ KE_{2\_initial} = \frac{1}{2} \times 1 \, \text{kg} \times (0 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{2\_initial} = 0 \, \text{J} \][/tex]
Total initial kinetic energy:
[tex]\[ KE_{initial} = KE_{1\_initial} + KE_{2\_initial} \][/tex]
[tex]\[ KE_{initial} = 24 \, \text{J} + 0 \, \text{J} \][/tex]
[tex]\[ KE_{initial} = 24 \, \text{J} \][/tex]
2. Calculate final kinetic energy:
Now we find the kinetic energy of each mass after the collision.
- For the 3 kg mass (now moving at 2 m/s):
[tex]\[ KE_{1\_final} = \frac{1}{2} \times 3 \, \text{kg} \times (2 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{1\_final} = \frac{1}{2} \times 3 \times 4 \][/tex]
[tex]\[ KE_{1\_final} = \frac{1}{2} \times 12 \][/tex]
[tex]\[ KE_{1\_final} = 6 \, \text{J} \][/tex]
- For the 1 kg mass (now moving at 6 m/s):
[tex]\[ KE_{2\_final} = \frac{1}{2} \times 1 \, \text{kg} \times (6 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{2\_final} = \frac{1}{2} \times 1 \times 36 \][/tex]
[tex]\[ KE_{2\_final} = \frac{1}{2} \times 36 \][/tex]
[tex]\[ KE_{2\_final} = 18 \, \text{J} \][/tex]
Total final kinetic energy:
[tex]\[ KE_{final} = KE_{1\_final} + KE_{2\_final} \][/tex]
[tex]\[ KE_{final} = 6 \, \text{J} + 18 \, \text{J} \][/tex]
[tex]\[ KE_{final} = 24 \, \text{J} \][/tex]
3. Check if kinetic energy is conserved:
[tex]\[ KE_{initial} = KE_{final} \][/tex]
[tex]\[ 24 \, \text{J} = 24 \, \text{J} \][/tex]
Since the initial and final kinetic energies are equal, the kinetic energy is conserved.
4. Calculate initial and final momentum:
The formula for momentum ([tex]\( p \)[/tex]) is:
[tex]\[ p = mv \][/tex]
where [tex]\( m \)[/tex] is the mass and [tex]\( v \)[/tex] is the velocity.
- Initial momentum:
[tex]\[ p_{initial} = (3 \, \text{kg} \times 4 \, \text{m/s}) + (1 \, \text{kg} \times 0 \, \text{m/s}) \][/tex]
[tex]\[ p_{initial} = 12 \, \text{kg} \cdot \text{m/s} \][/tex]
- Final momentum:
[tex]\[ p_{final} = (3 \, \text{kg} \times 2 \, \text{m/s}) + (1 \, \text{kg} \times 6 \, \text{m/s}) \][/tex]
[tex]\[ p_{final} = (3 \times 2) + (1 \times 6) \][/tex]
[tex]\[ p_{final} = 6 + 6 \][/tex]
[tex]\[ p_{final} = 12 \, \text{kg} \cdot \text{m/s} \][/tex]
5. Check if momentum is conserved:
[tex]\[ p_{initial} = p_{final} \][/tex]
[tex]\[ 12 \, \text{kg} \cdot \text{m/s} = 12 \, \text{kg} \cdot \text{m/s} \][/tex]
Since the initial and final momenta are equal, momentum is conserved.
### Conclusion:
Both kinetic energy and momentum are conserved in this collision. Therefore, the collision is elastic.
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