Nitha Mulungu
Sep 10
12000kg truck travelling at straight track at speed of 12m/s collide with identical stationary truck.If truck lock together after collision, calculate common speed and the loss of kinetic energy?
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ChatGPT
Sep 10
To solve this problem, we can use the principle of conservation of momentum and conservation of kinetic energy.
1. Conservation of momentum:
Before the collision:
Initial momentum = mass x velocity = 12000kg x 12m/s = 144000 kg m/s
After the collision:
Final momentum = (12000kg + 12000kg) x V (common speed)
Final momentum = 24000kg x V
According to the principle of conservation of momentum:
Initial momentum = Final momentum
144000 = 24000V
V = 6 m/s
Therefore, the common speed after the collision is 6 m/s.
2. Loss of kinetic energy:
Initial kinetic energy = 0.5 x mass x velocity^2
Initial kinetic energy = 0.5 x 12000kg x (12m/s)^2
Initial kinetic energy = 864000 J
Final kinetic energy = 0.5 x (12000kg + 12000kg) x (6m/s)^2
Final kinetic energy = 0.5 x 24000kg x 36
Final kinetic energy = 432000 J
Loss of kinetic energy = Initial kinetic energy - Final kinetic energy
Loss of kinetic energy = 864000 J - 432000 J
Loss of kinetic energy = 432000 J
Therefore, the loss of kinetic energy during the collision is 432000 Joules.
1. Conservation of momentum:
Before the collision:
Initial momentum = mass x velocity = 12000kg x 12m/s = 144000 kg m/s
After the collision:
Final momentum = (12000kg + 12000kg) x V (common speed)
Final momentum = 24000kg x V
According to the principle of conservation of momentum:
Initial momentum = Final momentum
144000 = 24000V
V = 6 m/s
Therefore, the common speed after the collision is 6 m/s.
2. Loss of kinetic energy:
Initial kinetic energy = 0.5 x mass x velocity^2
Initial kinetic energy = 0.5 x 12000kg x (12m/s)^2
Initial kinetic energy = 864000 J
Final kinetic energy = 0.5 x (12000kg + 12000kg) x (6m/s)^2
Final kinetic energy = 0.5 x 24000kg x 36
Final kinetic energy = 432000 J
Loss of kinetic energy = Initial kinetic energy - Final kinetic energy
Loss of kinetic energy = 864000 J - 432000 J
Loss of kinetic energy = 432000 J
Therefore, the loss of kinetic energy during the collision is 432000 Joules.
1
Google Gemini AI
Sep 10
Common Speed:
Momentum is conserved in the collision:
```
m1-v1 + m2-v2 = (m1 + m2)-vf
```
where:
- m1 and v1 are the mass and speed of the moving truck
- m2 and v2 are the mass and speed of the stationary truck (0 m/s)
- vf is the common speed after the collision
Substituting values:
```
12000 kg - 12 m/s + 12000 kg - 0 m/s = (12000 kg + 12000 kg) - vf
```
Solving for vf:
```
vf = 6 m/s
```
Therefore, the common speed of the trucks after the collision is 6 m/s.
Loss of Kinetic Energy:
Kinetic energy is not conserved in this collision because the trucks lock together and come to rest. The initial kinetic energy of the moving truck is:
```
KE1 = 1/2 - m1 - v1^2 = 1/2 - 12000 kg - (12 m/s)^2 = 864000 J
```
The final kinetic energy of the combined trucks is zero because they come to rest:
```
KE2 = 0 J
```
Therefore, the loss of kinetic energy is:
```
KE loss = KE1 - KE2 = 864000 J
```
Momentum is conserved in the collision:
```
m1-v1 + m2-v2 = (m1 + m2)-vf
```
where:
- m1 and v1 are the mass and speed of the moving truck
- m2 and v2 are the mass and speed of the stationary truck (0 m/s)
- vf is the common speed after the collision
Substituting values:
```
12000 kg - 12 m/s + 12000 kg - 0 m/s = (12000 kg + 12000 kg) - vf
```
Solving for vf:
```
vf = 6 m/s
```
Therefore, the common speed of the trucks after the collision is 6 m/s.
Loss of Kinetic Energy:
Kinetic energy is not conserved in this collision because the trucks lock together and come to rest. The initial kinetic energy of the moving truck is:
```
KE1 = 1/2 - m1 - v1^2 = 1/2 - 12000 kg - (12 m/s)^2 = 864000 J
```
The final kinetic energy of the combined trucks is zero because they come to rest:
```
KE2 = 0 J
```
Therefore, the loss of kinetic energy is:
```
KE loss = KE1 - KE2 = 864000 J
```
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