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Sagot :
Let's analyze the data for each trial to determine which one shows the conservation of momentum. Conservation of momentum states that the total initial momentum before the collision should equal the total final momentum after the collision.
### Trial 1
- Initial velocities:
- Ball A: [tex]\( +1.0 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -2.0 \, \text{m/s} \)[/tex]
- Final velocities:
- Ball A: [tex]\( -2.0 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -1.0 \, \text{m/s} \)[/tex]
Initial momentum:
[tex]\[ \text{Initial momentum} = (1.0 \, \text{kg} \times 1.0 \, \text{m/s}) + (1.0 \, \text{kg} \times -2.0 \, \text{m/s}) = 1.0 \, \text{kg.m/s} + (-2.0 \, \text{kg.m/s}) = -1.0 \, \text{kg.m/s} \][/tex]
Final momentum:
[tex]\[ \text{Final momentum} = (1.0 \, \text{kg} \times -2.0 \, \text{m/s}) + (1.0 \, \text{kg} \times -1.0 \, \text{m/s}) = (-2.0 \, \text{kg.m/s}) + (-1.0 \, \text{kg.m/s}) = -3.0 \, \text{kg.m/s} \][/tex]
Since the initial momentum is [tex]\(-1.0 \, \text{kg.m/s}\)[/tex] and the final momentum is [tex]\(-3.0 \, \text{kg.m/s}\)[/tex], momentum is not conserved in Trial 1.
### Trial 2
- Initial velocities:
- Ball A: [tex]\( +0.5 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -1.5 \, \text{m/s} \)[/tex]
- Final velocities:
- Ball A: [tex]\( -0.5 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -0.5 \, \text{m/s} \)[/tex]
Initial momentum:
[tex]\[ \text{Initial momentum} = (1.0 \, \text{kg} \times 0.5 \, \text{m/s}) + (1.0 \, \text{kg} \times -1.5 \, \text{m/s}) = 0.5 \, \text{kg.m/s} + (-1.5 \, \text{kg.m/s}) = -1.0 \, \text{kg.m/s} \][/tex]
Final momentum:
[tex]\[ \text{Final momentum} = (1.0 \, \text{kg} \times -0.5 \, \text{m/s}) + (1.0 \, \text{kg} \times -0.5 \, \text{m/s}) = (-0.5 \, \text{kg.m/s}) + (-0.5 \, \text{kg.m/s}) = -1.0 \, \text{kg.m/s} \][/tex]
Since the initial momentum is [tex]\(-1.0 \, \text{kg.m/s}\)[/tex] and the final momentum is [tex]\(-1.0 \, \text{kg.m/s}\)[/tex], momentum is conserved in Trial 2.
### Trial 3
- Initial velocities:
- Ball A: [tex]\( +2.0 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( +1.0 \, \text{m/s} \)[/tex]
- Final velocities:
- Ball A: [tex]\( +1.0 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -2.0 \, \text{m/s} \)[/tex]
Initial momentum:
[tex]\[ \text{Initial momentum} = (1.0 \, \text{kg} \times 2.0 \, \text{m/s}) + (1.0 \, \text{kg} \times 1.0 \, \text{m/s}) = 2.0 \, \text{kg.m/s} + 1.0 \, \text{kg.m/s} = 3.0 \, \text{kg.m/s} \][/tex]
Final momentum:
[tex]\[ \text{Final momentum} = (1.0 \, \text{kg} \times 1.0 \, \text{m/s}) + (1.0 \, \text{kg} \times -2.0 \, \text{m/s}) = 1.0 \, \text{kg.m/s} + (-2.0 \, \text{kg.m/s}) = -1.0 \, \text{kg.m/s} \][/tex]
Since the initial momentum is [tex]\(3.0 \, \text{kg.m/s}\)[/tex] and the final momentum is [tex]\(-1.0 \, \text{kg.m/s}\)[/tex], momentum is not conserved in Trial 3.
### Trial 4
- Initial velocities:
- Ball A: [tex]\( +0.5 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -1.0 \, \text{m/s} \)[/tex]
- Final velocities:
- Ball A: [tex]\( +1.5 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -1.5 \, \text{m/s} \)[/tex]
Initial momentum:
[tex]\[ \text{Initial momentum} = (1.0 \, \text{kg} \times 0.5 \, \text{m/s}) + (1.0 \, \text{kg} \times -1.0 \, \text{m/s}) = 0.5 \, \text{kg.m/s} + (-1.0 \, \text{kg.m/s}) = -0.5 \, \text{kg.m/s} \][/tex]
Final momentum:
[tex]\[ \text{Final momentum} = (1.0 \, \text{kg} \times 1.5 \, \text{m/s}) + (1.0 \, \text{kg} \times -1.5 \, \text{m/s}) = 1.5 \, \text{kg.m/s} + (-1.5 \, \text{kg.m/s}) = 0.0 \, \text{kg.m/s} \][/tex]
Since the initial momentum is [tex]\(-0.5 \, \text{kg.m/s}\)[/tex] and the final momentum is [tex]\(0.0 \, \text{kg.m/s}\)[/tex], momentum is not conserved in Trial 4.
### Conclusion
Analyzing the results, we find that:
- Trial 1: Momentum is not conserved.
- Trial 2: Momentum is conserved.
- Trial 3: Momentum is not conserved.
- Trial 4: Momentum is not conserved.
Thus, Trial 2 shows the conservation of momentum in a closed system.
### Trial 1
- Initial velocities:
- Ball A: [tex]\( +1.0 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -2.0 \, \text{m/s} \)[/tex]
- Final velocities:
- Ball A: [tex]\( -2.0 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -1.0 \, \text{m/s} \)[/tex]
Initial momentum:
[tex]\[ \text{Initial momentum} = (1.0 \, \text{kg} \times 1.0 \, \text{m/s}) + (1.0 \, \text{kg} \times -2.0 \, \text{m/s}) = 1.0 \, \text{kg.m/s} + (-2.0 \, \text{kg.m/s}) = -1.0 \, \text{kg.m/s} \][/tex]
Final momentum:
[tex]\[ \text{Final momentum} = (1.0 \, \text{kg} \times -2.0 \, \text{m/s}) + (1.0 \, \text{kg} \times -1.0 \, \text{m/s}) = (-2.0 \, \text{kg.m/s}) + (-1.0 \, \text{kg.m/s}) = -3.0 \, \text{kg.m/s} \][/tex]
Since the initial momentum is [tex]\(-1.0 \, \text{kg.m/s}\)[/tex] and the final momentum is [tex]\(-3.0 \, \text{kg.m/s}\)[/tex], momentum is not conserved in Trial 1.
### Trial 2
- Initial velocities:
- Ball A: [tex]\( +0.5 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -1.5 \, \text{m/s} \)[/tex]
- Final velocities:
- Ball A: [tex]\( -0.5 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -0.5 \, \text{m/s} \)[/tex]
Initial momentum:
[tex]\[ \text{Initial momentum} = (1.0 \, \text{kg} \times 0.5 \, \text{m/s}) + (1.0 \, \text{kg} \times -1.5 \, \text{m/s}) = 0.5 \, \text{kg.m/s} + (-1.5 \, \text{kg.m/s}) = -1.0 \, \text{kg.m/s} \][/tex]
Final momentum:
[tex]\[ \text{Final momentum} = (1.0 \, \text{kg} \times -0.5 \, \text{m/s}) + (1.0 \, \text{kg} \times -0.5 \, \text{m/s}) = (-0.5 \, \text{kg.m/s}) + (-0.5 \, \text{kg.m/s}) = -1.0 \, \text{kg.m/s} \][/tex]
Since the initial momentum is [tex]\(-1.0 \, \text{kg.m/s}\)[/tex] and the final momentum is [tex]\(-1.0 \, \text{kg.m/s}\)[/tex], momentum is conserved in Trial 2.
### Trial 3
- Initial velocities:
- Ball A: [tex]\( +2.0 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( +1.0 \, \text{m/s} \)[/tex]
- Final velocities:
- Ball A: [tex]\( +1.0 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -2.0 \, \text{m/s} \)[/tex]
Initial momentum:
[tex]\[ \text{Initial momentum} = (1.0 \, \text{kg} \times 2.0 \, \text{m/s}) + (1.0 \, \text{kg} \times 1.0 \, \text{m/s}) = 2.0 \, \text{kg.m/s} + 1.0 \, \text{kg.m/s} = 3.0 \, \text{kg.m/s} \][/tex]
Final momentum:
[tex]\[ \text{Final momentum} = (1.0 \, \text{kg} \times 1.0 \, \text{m/s}) + (1.0 \, \text{kg} \times -2.0 \, \text{m/s}) = 1.0 \, \text{kg.m/s} + (-2.0 \, \text{kg.m/s}) = -1.0 \, \text{kg.m/s} \][/tex]
Since the initial momentum is [tex]\(3.0 \, \text{kg.m/s}\)[/tex] and the final momentum is [tex]\(-1.0 \, \text{kg.m/s}\)[/tex], momentum is not conserved in Trial 3.
### Trial 4
- Initial velocities:
- Ball A: [tex]\( +0.5 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -1.0 \, \text{m/s} \)[/tex]
- Final velocities:
- Ball A: [tex]\( +1.5 \, \text{m/s} \)[/tex]
- Ball B: [tex]\( -1.5 \, \text{m/s} \)[/tex]
Initial momentum:
[tex]\[ \text{Initial momentum} = (1.0 \, \text{kg} \times 0.5 \, \text{m/s}) + (1.0 \, \text{kg} \times -1.0 \, \text{m/s}) = 0.5 \, \text{kg.m/s} + (-1.0 \, \text{kg.m/s}) = -0.5 \, \text{kg.m/s} \][/tex]
Final momentum:
[tex]\[ \text{Final momentum} = (1.0 \, \text{kg} \times 1.5 \, \text{m/s}) + (1.0 \, \text{kg} \times -1.5 \, \text{m/s}) = 1.5 \, \text{kg.m/s} + (-1.5 \, \text{kg.m/s}) = 0.0 \, \text{kg.m/s} \][/tex]
Since the initial momentum is [tex]\(-0.5 \, \text{kg.m/s}\)[/tex] and the final momentum is [tex]\(0.0 \, \text{kg.m/s}\)[/tex], momentum is not conserved in Trial 4.
### Conclusion
Analyzing the results, we find that:
- Trial 1: Momentum is not conserved.
- Trial 2: Momentum is conserved.
- Trial 3: Momentum is not conserved.
- Trial 4: Momentum is not conserved.
Thus, Trial 2 shows the conservation of momentum in a closed system.
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