To analyze the situation where a lab cart of mass [tex]\(15 \, \text{kg}\)[/tex] is moving with a constant velocity [tex]\(v_i\)[/tex] and a [tex]\(2 \, \text{kg}\)[/tex] mass is dropped into it from directly above, let's use the principle of conservation of momentum. Here is a step-by-step solution:
1. Identify the Initial Momentum:
- The initial momentum of the system is purely due to the moving cart, as the [tex]\(2 \, \text{kg}\)[/tex] mass is just dropped directly into the cart from rest.
- The cart's initial momentum:
[tex]\[ P_{\text{cart, initial}} = 15 \, \text{kg} \cdot v_i \][/tex]
- Since the [tex]\(2 \, \text{kg}\)[/tex] mass has no initial horizontal velocity, its initial momentum is zero:
[tex]\[ P_{\text{mass, initial}} = 2 \, \text{kg} \cdot 0 = 0 \][/tex]
- Therefore, the total initial momentum [tex]\(P_{\text{initial}}\)[/tex] is:
[tex]\[ P_{\text{initial}} = 15 \, \text{kg} \cdot v_i + 2 \, \text{kg} \cdot 0 \][/tex]
[tex]\[ P_{\text{initial}} = 15 v_i \][/tex]
2. Identify the Final Momentum:
- After the mass is dropped into the cart, the total mass of the cart system becomes:
[tex]\[ \text{Total mass} = 15 \, \text{kg} + 2 \, \text{kg} = 17 \, \text{kg} \][/tex]
- Let [tex]\(v_f\)[/tex] be the final velocity of the cart system after the mass is dropped.
- The final momentum of the system:
[tex]\[ P_{\text{final}} = \text{Total mass} \cdot v_f = 17 \, \text{kg} \cdot v_f \][/tex]
3. Apply the Conservation of Momentum:
- According to the law of conservation of momentum, the total initial momentum must equal the total final momentum:
[tex]\[ P_{\text{initial}} = P_{\text{final}} \][/tex]
[tex]\[ 15 v_i = 17 v_f \][/tex]
Given these steps, we can now evaluate the provided options to find the equation that best represents the horizontal momentum:
[tex]\[
\begin{align*}
\text{Option 1:}\quad & 15 v_i + 2 v_i = 15(0) + 2 v_i \quad \implies \text{Incorrect (initial mass terms do not match)} \\
\text{Option 2:}\quad & 15 v_i + 2(0) = 15 v_t + 2(0) \quad \implies \text{Incorrect (missing total mass change)} \\
\text{Option 3:}\quad & 15 v_i + 2(0) = (15 + 2) v_f \quad \implies \text{Correct (conservation of momentum correctly stated)} \\
\text{Option 4:}\quad & 15(0) + 2 v_i = (15 + 2) v_t \quad \implies \text{Incorrect (initial state does not match scenario)}
\end{align*}
\][/tex]
Therefore, the equation that best represents the horizontal momentum in this situation is:
[tex]\[15 v_i + 2(0) = (15 + 2) v_f\][/tex]
So, the correct answer is:
[tex]\[ \boxed{3} \][/tex]
