To solve the problem of horizontal momentum conservation when a mass is dropped into a moving cart, we have to apply the principle of conservation of momentum. Let's break it down step-by-step:
1. Initial Conditions:
- Mass of the cart, [tex]\( m_c = 15 \)[/tex] kg.
- Initial velocity of the cart, [tex]\( v_i \)[/tex].
- Mass being dropped, [tex]\( m_d = 2 \)[/tex] kg.
- The mass is dropped vertically, so its initial horizontal velocity is 0.
2. Momentum Before Mass is Dropped:
- The only moving object initially is the cart.
- Initial momentum, [tex]\( P_i \)[/tex], is given by the product of the mass of the cart and its velocity.
[tex]\[
P_i = m_c \cdot v_i = 15 \cdot v_i
\][/tex]
- The dropped mass does not add to the initial horizontal momentum because its initial horizontal velocity is 0.
3. Momentum After Mass is Dropped:
- After the mass is dropped into the cart, the cart and the mass move together with a common final velocity, [tex]\( v_f \)[/tex].
- The combined mass after the drop is [tex]\( m_c + m_d = 15 + 2 = 17 \)[/tex] kg.
- The final momentum, [tex]\( P_f \)[/tex], is given by the product of the combined mass and the final velocity.
[tex]\[
P_f = (m_c + m_d) \cdot v_f = 17 \cdot v_f
\][/tex]
4. Conservation of Momentum:
- According to the conservation of momentum, the total initial momentum [tex]\( P_i \)[/tex] must equal the total final momentum [tex]\( P_f \)[/tex]:
[tex]\[
P_i = P_f
\][/tex]
Substituting the values we get:
[tex]\[
15 \cdot v_i = 17 \cdot v_f
\][/tex]
Therefore, matching with the given options, 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 choice is:
[tex]\[
\boxed{15 v_i + 2(0) = (15 + 2) v_f}
\][/tex]
