Problem 66 - Master Resource Book in JEE Main Physics

Question

A bomb of mass \(9 \mathrm{~kg}\) explodes into 2 pieces of mass \(3 \mathrm{~kg}\) and \(6 \mathrm{~kg}\). The velocity of mass \(3 \mathrm{~kg}\) is \(1.6 \mathrm{~m} / \mathrm{s}\), the kinetic energy of mass \(6 \mathrm{~kg}\) is (a) \(3.84 \mathrm{~J}\) (b) \(9.6 \mathrm{~J}\) (c) \(1.92 \mathrm{~J}\) (d) \(2.92 \mathrm{~J}\)

Step-by-Step Solution

Answer

The kinetic energy of the second piece is \( 1.92 \text{ J} \).

Step 1: Understand the Conservation of Momentum

When the bomb explodes, the total momentum before and after the explosion remains constant. Initially, the bomb is at rest, so the total initial momentum is zero. Thus, the final momentum of the system must also be zero. This can be expressed as: \( m_1 \cdot v_1 + m_2 \cdot v_2 = 0 \), where \( m_1 = 3 \text{ kg} \), \( v_1 = 1.6 \text{ m/s} \), and \( v_2 \) is the velocity of the second piece.

Step 2: Calculate Velocity of Second Piece

Substitute the known values into the momentum equation to find \( v_2 \): \[ 3 \cdot 1.6 + 6 \cdot v_2 = 0 \] Solving for \( v_2 \) gives: \[ v_2 = -\frac{3 \cdot 1.6}{6} = -0.8 \text{ m/s} \] The negative sign indicates that the direction of the second piece's velocity is opposite to that of the first piece.

Step 3: Determine Kinetic Energy of Second Piece

Kinetic energy is given by the formula: \[ KE = \frac{1}{2} m v^2 \] Substitute the mass and velocity of the second piece into this formula: \[ KE = \frac{1}{2} \times 6 \times (-0.8)^2 = \frac{1}{2} \times 6 \times 0.64 = 1.92 \text{ J} \] Thus, the kinetic energy of the second piece is \( 1.92 \text{ J} \).

Key Concepts

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. It can be calculated using the formula: \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the object.
In the context of explosions, like the one described in the exercise, each fragment of the exploded mass will have its own kinetic energy based on its mass and velocity.

• The kinetic energy of each piece is determined independently.
• The total kinetic energy in the system may increase, even though momentum remains constant.

In this specific problem, each fragment's kinetic energy depends on their respective masses and velocities post-explosion.
The smaller piece moves at a velocity of \( 1.6 \text{ m/s} \), while the larger piece moves at \(-0.8 \text{ m/s}\), which is crucial for determining the kinetic energy for each piece.

Physics Problem Solving

Physics problem solving often requires applying fundamental concepts like momentum and energy conservation laws.
Understanding these concepts and how to use them is key to solving physics problems correctly.

• First, identify the principles at play, such as conservation of momentum in this explosion problem.
• Set up equations based on these principles to represent the given situation.
• Carefully substitute known values.
• Solve the equations to find unknown quantities.

In the given problem, the conservation of momentum equation helped find the velocity of the second piece.
This velocity could then be used to determine its kinetic energy. These steps highlight the step-by-step approach in physics problem solving, where breaking down the problem into manageable parts is critical.

Explosion Mechanics

Explosion mechanics involves studying the behavior of materials that burst into pieces, transferring matter and energy rapidly.
Such events are often governed by principles like conservation of momentum and energy changes.

• An explosion starts with an object at rest and ends with the distribution of its fragments.
• Conservation of momentum assures that the momentum before and after the explosion stay the same.
• Energy transformation occurs, where the chemical potential energy is converted into kinetic energy.

In the problem, the bomb's explosion causes it to separate into two parts with differing masses and velocities.
Understanding the mechanics of how velocities change post-explosion allows us to calculate individual kinetic energies, even as the system follows the conservation rules.