A ball is dropped from a height of 10 m . Ball is embedded in sand through 1 m and stops. Then why is momentum conserved .?

When we talk about momentum conservation, it's essential to understand the system we're examining and the forces at play. In the scenario you've described, a ball is dropped from a height of 10 meters, and upon hitting the ground, it becomes embedded in sand after penetrating 1 meter. The question of why momentum is conserved in this situation can be addressed by analyzing the interactions involved.

Understanding Momentum Conservation

Momentum is defined as the product of an object's mass and its velocity. In a closed system where no external forces act, the total momentum before an event must equal the total momentum after that event. However, in real-world situations, we often have to consider external forces, such as gravity and friction, which can complicate our analysis.

Breaking Down the Scenario

In this case, let's break down the events:

• The ball is dropped from a height of 10 meters, gaining speed as it falls due to gravity.
• Upon impact with the sand, the ball transfers its momentum to the sand as it becomes embedded.

Initially, the ball has momentum as it falls. When it hits the sand, it exerts a force on the sand, and in response, the sand exerts an equal and opposite force on the ball (as per Newton's third law). This interaction is crucial for understanding momentum conservation.

Analyzing the Forces

While the ball is falling, the only significant external force acting on it is gravity. When the ball hits the sand, it decelerates rapidly, and its momentum decreases. However, the sand also gains momentum in the opposite direction. The key point here is that the system includes both the ball and the sand. When we consider the momentum of both the ball and the sand together, we find that the total momentum before and after the impact remains constant.

Calculating Momentum Changes

To illustrate this with numbers, let's assume the mass of the ball is 0.5 kg. When dropped from 10 m, we can calculate its velocity just before impact using the equation:

v = √(2gh)

Where:

• g = 9.81 m/s² (acceleration due to gravity)
• h = 10 m (height)

Calculating this gives:

v = √(2 * 9.81 * 10) ≈ 14 m/s

Now, the momentum of the ball just before it hits the sand is:

p_ball = mass * velocity = 0.5 kg * 14 m/s = 7 kg·m/s

Upon embedding in the sand, the ball comes to a stop, so its final momentum is 0 kg·m/s. However, the sand gains momentum equal to the momentum lost by the ball. If we consider a small volume of sand that moves with the ball, it will gain momentum equal to 7 kg·m/s in the opposite direction. Thus, the total momentum of the system (ball + sand) remains conserved.

Final Thoughts

In summary, while the ball loses momentum as it stops, the sand gains an equal amount of momentum, ensuring that the total momentum of the system remains unchanged. This principle of conservation of momentum is a fundamental concept in physics, illustrating how interactions between objects can lead to a balance in momentum, even when one object comes to a stop. Understanding these interactions helps clarify why momentum conservation holds true in this scenario.