A hot-air balloon is rising upwards at a constant velocity of 5m.s . When the balloon is 100m above the ground , a sandbag is dropped from it. Will the velocity of the hot-air balloon increase , decrease or remain the same immediately after the sandbag has been released ? Explain fully for 5 marks.

The velocity will immediately begin to accelerate upward, but it will begin at its original value (5 m/s)

5 m/s

When the sandbag is dropped from the hot-air balloon, the velocity of the balloon will remain the same. Here's a step-by-step explanation:

1. The hot-air balloon is initially rising upwards at a constant velocity of 5m/s. This means that it is moving upwards at a steady speed of 5 meters per second.

2. When the sandbag is dropped, two things happen: gravity acts on the sandbag and pulls it downward, and the hot-air balloon continues to move upwards with the same velocity.

3. According to Newton's first law of motion, an object will continue to move with the same velocity unless acted upon by an external force. In this case, the hot-air balloon is not experiencing any external force that would change its upward velocity.

4. The sandbag, on the other hand, will be affected by the force of gravity and will accelerate downward. However, this acceleration does not affect the upward velocity of the balloon.

5. As a result, the velocity of the hot-air balloon will remain the same immediately after the sandbag has been released. The upward velocity of 5m/s will not increase or decrease.

To summarize, the velocity of the hot-air balloon will remain constant because, in the absence of any external forces, it will continue to move upward with the same velocity even after the sandbag is dropped.

To determine the effect of dropping the sandbag on the velocity of the hot-air balloon, we need to consider the principle of conservation of momentum.

The momentum of an object is defined as the product of its mass and velocity. In this case, the sandbag has mass, so its change in momentum will affect the momentum of the hot-air balloon.

When the sandbag is dropped, it falls freely due to the force of gravity. Neglecting air resistance, it will accelerate downward at a rate of approximately 9.8 m/s². Since the sandbag is no longer part of the balloon system, we can apply the principle of conservation of momentum to analyze the impact of this event on the balloon's velocity.

Conservation of momentum states that the total momentum of an isolated system remains constant if no external forces are acting on it. In this case, the hot-air balloon and the sandbag were originally in a state of equilibrium since the balloon was rising at a constant velocity.

However, when the sandbag is dropped, an external force (gravity) acts on it, causing it to accelerate downward. As the sandbag gains momentum in the downward direction, the total momentum of the system (balloon + sandbag) changes. Since there are no external forces acting on the balloon itself, the momentum of the balloon must remain constant.

Therefore, the velocity of the hot-air balloon will remain the same immediately after the sandbag has been released. The additional downward momentum gained by the sandbag is canceled out by the unchanged momentum of the balloon, resulting in no change in velocity.