Linear Momentum in a Lab Cart Scenario
What equation best represents the horizontal momentum in a lab cart scenario where a 2 kg mass is dropped into a 15 kg moving cart?
A. 15vi 2vi = 15vf 2vf
B. 15vi 2(0) = 15vf 2vf
C. 15vi 2(0) = (15 + 2)vf
D. 15vi 2vi = (15 + 2)vf
Answer:
The equation [tex]15v_{i} + 2*0 = (15 + 2)v_{f} [/tex] (option C) 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 in this scenario can be calculated using the principle of conservation of linear momentum. Let's break it down:
Given parameters: m₁: mass of the lab cart = 15 kg m₂: mass of the object dropped = 2 kg v₁ᵢ: initial velocity of the lab cart v₂ᵢ: initial velocity of the object (dropped, so initially at rest) v₁f: final velocity of the lab cart v₂f: final velocity of the object
According to the conservation of linear momentum: [tex] m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f} [/tex]
Substitute the given values: [tex] 15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f} [/tex]
Since the final velocity of the lab cart and the object will be the same after the collision, we can simplify the equation to: [tex] 15v_{1}_{i} + 2*0 = v_{f}(15 + 2) [/tex]
Therefore, the correct equation representing the horizontal momentum in this scenario is: [tex]15v_{i} + 2*0 = (15 + 2)v_{f} [/tex]