Work = (force) x (distance) = (450 N) x (4 m) = 1,800 joules
Power = (work) / (time) = (1,800 joules) / (2 sec) = 900 watts .
Chemical energy. Elimination is useful to answer this question.
Answer:
4 kg → +4 m/s
5 kg → -5 m/s
Explanation:
The law of conservation of momentum states that:
- m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
- left side → velocities before collision
- right side → velocities after collision
You'll notice that we have two missing variables: v₁' & v₂'. Assuming this is a perfectly elastic collision, we can use the conservation of kinetic energy to set the initial and final velocities of the individual bodies equal to each other.
Let's substitute all known variables into the first equation.
- (4)(-6) + (5)(3) = (4)v₁' + (5)v₂'
- -24 + 15 = 4v₁' + 5v₂'
- -9 = 4v₁' + 5v₂'
Let's substitute the known variables into the second equation.
- (-6) + v₁' = (3) + v₂'
- -9 = -v₁' + v₂'
- 9 = v₁' - v₂'
Now we have a system of equations where we can solve for v₁ and v₂.
- -9 = 4v₁' + 5v₂'
- 9 = v₁' - v₂'
Use the elimination method and multiply the bottom equation by -4.
- -9 = 4v₁' + 5v₂'
- -36 = -4v₁' + 4v₂'
Add the equations together.
<u>The final velocity of the second body (5 kg) is -5 m/s</u>. Substitute this value into one of the equations in the system to find v₁.
- 9 = v₁' - v₂'
- 9 = v₁' - (-5)
- 9 = v₁' + 5
- 4 = v₁'
<u>The final velocity of the first body (4 kg) is 4 m/s.</u>
<u></u>
We can verify our answer by making sure that the law of conservation of momentum is followed.
- m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
- (4)(-6) + (5)(3) = (4)(4) + (5)(-5)
- -24 + 15 = 16 - 25
- -9 = -9
The combined momentum of the bodies before the collision is equal to the combined momentum of the bodies after the collision. [✓]
