A 2.8 kg cart is moving horizontally at an initial speed of 6.6 m/s across a frictionless surface when a 8.6 kg load is dropped vertically onto it. The vertical speed of the load is 4.1 m/s just prior to impact. What is the final horizontal speed of the combined cart and load after the impact?

Question

BlackTom AI · Accepted Answer

To solve this problem, we apply the principle of conservation of linear momentum in the horizontal direction. Since the load is dropped vertically, its initial horizontal velocity is zero, and it does not affect the horizontal momentum before the impact.

The initial horizontal momentum of the system is given by the momentum of the cart alone, as the load has no horizontal component of velocity:

$$ p_{initial} = m_{cart} \times v_{cart} = 2.8 \, \text{kg} \times 6.6 \, \text{m/s} = 18.48 \, \text{kg} \cdot \text{m/s} $$

After the load is dropped onto the cart, the total mass of the system becomes $ m_{total} = m_{cart} + m_{load} = 2.8 \, \text{kg} + 8.6 \, \text{kg} = 11.4 \, \text{kg} $.

Let $ v_{final} $ be the final horizontal speed of the combined system. By conservation of momentum:

$$ p_{initial} = p_{final} $$
$$ 18.48 \, \text{kg} \cdot \text{m/s} = 11.4 \, \text{kg} \times v_{final} $$

Solving for $ v_{final} $:

$$ v_{final} = \frac{18.48 \, \text{kg} \cdot \text{m/s}}{11.4 \, \text{kg}} = 1.62 \, \text{m/s} $$

Thus, the final horizontal speed of the combined cart and load is $ 1.62 \, \text{m/s} $. Therefore, the correct answer is option a. 1.62 m/s. Other options do not satisfy the conservation of momentum in the horizontal direction.

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