The equation (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.
The horizontal momentum is given by:
Where:
- m₁: is the mass of the lab cart = 15 kg
- m₂: is the <em>mass </em>of the object dropped = 2 kg
- : is the initial velocity of the<em> lab cart </em>
- : is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
- : is the final velocity of the<em> lab cart </em>
- : is the <em>final velocity</em> of the <em>object </em>
Then, the horizontal momentum is:
When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:
Therefore, the equation represents the horizontal momentum (option 3).
Learn more about linear momentum here:
I hope it helps you!
Yes, this is correct Answer.
Linear momentum has to be conserved. It was zero before the thread eas burned ... when nothing was moving ... so the momentum of the masses moving in opposite directions has to add up to zero. ... Momentum = mass times speed. ... In one direction, you have 5 kg times 1/5 m/s= 1 kg-m/s. ... We need 1 kg-m/s in the other direction. ... 7 kg times speed = 1 kg-m/s. ... Can you finish it from here ?
The new speed of car is 10.9 m/s
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According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.
Mass of the railroad car, m₁ = 7950 kg
Mass of the load, m₂ = 2950 kg
It can be assumed as the speed of the car, u₁ = 15 m/s
Initially, it is at rest, u₂ = 0
Let v is the speed of the car. It can be calculated using the conservation of momentum as :
Therefore, the new speed of care is 10.9 m/s
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