Problem 473 - JEE Physics

Question

A sphere of mass \(\mathrm{m}\) moving the velocity \(\mathrm{v}\) enters a hanging bag of sand and stops. If the mass of the bag is \(\mathrm{M}\) and it is raised by height \(\mathrm{h}\), then the velocity of the sphere was (A) \([\\{\mathrm{m}+\mathrm{M}\\} / \mathrm{m}] \sqrt{(2 \mathrm{gh})}\) (B) \((\mathrm{M} / \mathrm{m}) \sqrt{(} 2 \mathrm{gh})\) (C) \([\mathrm{m} /\\{\mathrm{M}+\mathrm{m}\\}] \sqrt{(2 \mathrm{gh})}\) (D) \((\mathrm{m} / \mathrm{M}) \sqrt{(2 \mathrm{gh})}\)

Step-by-Step Solution

Answer

The short answer is: \(v=\frac{M + m}{m}\sqrt{2gh}\).

Step 1: Applying the conservation of momentum

Before the collision, the momentum of the sphere is \(m \cdot v\) and the momentum of the sandbag is 0. After the collision, the momentum of the sphere is 0 and the momentum of the sandbag is \((M + m) \cdot v_{f}\). According to the conservation of momentum principle: \[m \cdot v = (M + m) \cdot v_{f}\] Now, we need to find the final velocity \(v_{f}\) of the sandbag and sphere after the collision.

Step 2: Applying the conservation of energy

After the collision, the sphere and sandbag move together at final velocity \(v_{f}\) and rise to a height h. The potential energy at height h is \((M+m)gh\). Due to the conservation of energy principle: \[\frac{1}{2}(M+m)v_{f}^2 = (M+m)gh\] Next, we need to isolate \(v_f\) from the equation above.

Step 3: Solving for the final velocity

To find the final velocity \(v_{f}\), divide both sides of the energy conservation equation by \((M + m)\): \[\frac{1}{2}v_{f}^2 = gh\] Now, take the square root of both sides to get \(v_{f}\): \[v_{f} = \sqrt{2gh}\]

Step 4: Solving for the initial velocity of the sphere

Substitute \(v_{f}\) from Step 3 into the conservation of momentum equation from Step 1: \[m \cdot v = (M + m) \cdot \sqrt{2gh}\] Finally, solve for the initial velocity v of the sphere: \[v=\frac{M + m}{m}\sqrt{2gh}\] The correct answer corresponds to the option (A).

Key Concepts

Conservation of Energy

When we talk about the conservation of energy, we're essentially discussing how energy is transferred or transformed, but not lost. In the exercise above, this concept plays a major role.
Imagine energy as a fixed amount of something valuable. You can change its form, but you can't make it disappear. That's what the conservation of energy tells us. In this scenario, when the sphere collides with the sandbag, the kinetic energy it had is transferred and changes into potential energy as the sandbag rises to a height.
So, at the start, all the energy was kinetic because the sphere was in motion. After the collision and when the bag moves upward, that energy now becomes potential. This principle can be expressed in the equation:

• Kinetic Energy: \ \(\frac{1}{2}(M+m)v_{f}^2 \ \)
• Potential Energy: \ \((M+m)gh \ \)

In this example, the kinetic energy right after the collision is converted fully into the potential energy when the sandbag reaches its highest point. By using these expressions and understanding how they relate, we can find the values we need to solve problems in physics.

Collision Problems in Physics

Collisions are all about objects interacting, often transforming the motion and energy from one to another. Here, we consider the sphere hitting the sandbag as a classic collision problem.
Think of a collision as a way to exchange momentum and energy. When the sphere enters the sandbag, it's a collision that obeys laws like the conservation of momentum and energy. This involves:

• Before the Collision: The sphere's momentum is determined by its mass and velocity \((m \cdot v)\)
• After the Collision: The combined system's momentum \((M + m) \cdot v_{f}\)

According to the conservation of momentum, no momentum is lost; it's simply transferred. Knowing this helps us calculate the motion and speed of objects post-collision.
By rearranging the momentum equation, we see how the initial velocity of the sphere affects the sandbag, demonstrating both the interconnected and predictable nature of collisions in physics.

Potential and Kinetic Energy

The concept of potential and kinetic energy is foundational in physics, helping us understand motion and position-related energy.
Kinetic energy, related to movement, is present when the sphere is in motion. This can be calculated using the formula:

• Kinetic Energy (\(KE\)): \ \(\frac{1}{2}mv^2 \)

Potential energy, on the other side, is about the object's position. When the sandbag rises to a certain height, it's storing energy due to gravity:

• Potential Energy (\(PE\)): \ \(mgh \)

The switch from kinetic to potential energy in this exercise demonstrates these two forms of energy at work. Initially, the sphere's kinetic energy is turned into the potential energy of the sandbag after the collision.
Understanding how objects convert kinetic to potential energy, and vice versa, gives insight into how energies interact in physical systems. This can be fundamental to solving more complex physics problems.