Question

A 7.0-kg shell at rest explodes into two fragments, one with a mass of 2.0 kg and the other with a mass of 5.0 kg. If the heavier fragment gains 100 J of kinetic energy from the explosion, how much kinetic energy does the lighter one gain?

Short answer

The lighter fragment gains 250 J of kinetic energy.

Step-by-step solution

Understand the Law of Conservation of Momentum

The initial momentum of the system is zero since the shell is at rest. According to the law of conservation of momentum, the total momentum after the explosion should also be zero.

Assign Variables and Formulate the Momentum Equation

Let the velocity of the 2 kg fragment be \( v_1 \) and the velocity of the 5 kg fragment be \( v_2 \). The momentum equation can be written as: \( 2v_1 + 5v_2 = 0 \). Solve for one of the variables in terms of the other, i.e., \( v_1 = -\frac{5}{2}v_2 \).

Calculate the Velocity of the Heavier Fragment Using Kinetic Energy

The kinetic energy of the 5 kg fragment is given by \( KE_{5} = \frac{1}{2} \times 5 \times v_2^2 = 100 \text{ J} \). Solving for \( v_2 \) gives: \( v_2 = \sqrt{\frac{200}{5}} = \sqrt{40} \).

Calculate the Velocity of the Lighter Fragment

Substitute \( v_2 = \sqrt{40} \) into the equation \( v_1 = -\frac{5}{2}v_2 \). Therefore, \( v_1 = -\frac{5}{2}\sqrt{40} \).

Compute the Kinetic Energy of the Lighter Fragment

The kinetic energy of the 2 kg fragment is \( KE_{2} = \frac{1}{2} \times 2 \times v_1^2 \). Substitute \( v_1 \) to find \( KE_{2} = \frac{1}{2} \times 2 \times \left(-\frac{5}{2}\sqrt{40}\right)^2 \). Simplifying gives \( KE_{2} = 250 \text{ J} \).

Key concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses because of its motion. When an object is moving, it holds kinetic energy, which can be calculated using the formula:
\[KE = \frac{1}{2}mv^2\]where \( m \) is the mass of the object and \( v \) is its velocity.
In the context of the exploding shell, we know that kinetic energy is transferred to the fragments as they move away from each other. The heavier fragment of mass 5.0 kg gains 100 J of kinetic energy. By using the formula, we can see how velocity is tied to the energy gained. Solving for velocity helps determine the kinetic energy of the other fragment. It is important to remember this relationship when calculating the effects of an explosion.
Quick pointers to remember about kinetic energy:

• Kinetic energy is directly proportional to the mass and the square of velocity.
• When velocity doubles, kinetic energy increases fourfold.
• The unit of measurement for kinetic energy is joules (J).

Explosion Mechanics

Explosion mechanics deals with how energy is distributed during an explosion and how it affects the motion of fragments. An explosion is a sudden and often violent release of energy (like in the shell), leading to fragments or particles being separated with high speeds.
In our exercise, a single explosion resulted in the separation of a fragmented shell into two parts. The conservation of momentum plays a huge role here: Before the explosion, the shell is at rest, and its total momentum is zero. After the explosion, despite the fragments flying apart, the total system's momentum remains zero.
The energy distribution in an explosion is also crucial. Kinetic energy is released from stored potential energy, causing fragments to spread out at different velocities. The heavier fragment in the exercise received a set amount of kinetic energy, allowing the calculations of the lighter fragment's energy.
Understanding explosion mechanics can help solve many physics problems related to motion and energy transfer. Key points in explosion mechanics:

• Explosions convert stored energy into kinetic energy of fragments.
• Momentum is conserved at all times, even during an explosion.
• Energy distribution affects how fast and in which direction fragments move.

Physics Problem Solving

Solving physics problems involves a systematic approach to simplify and handle various parameters involved. For the given problem of a shell exploding, a step-by-step approach improves understanding and accuracy.
Here are effective strategies used to solve such problems:

• Understand the principles involved: Initially, identify key physics laws such as conservation of momentum and energy conservation. They form the basis for solving the problem.
• Define variables clearly: Assign variables such as mass and velocity to different elements, and make sure to use units consistently.
• Simplify with equations: Use known formulas to express relationships. The momentum equation \( 2v_1 + 5v_2 = 0 \) was crucial here.
• Calculate step-by-step: Break the problem into smaller steps, such as calculating individual velocities from given kinetic energy.
• Verify your results: Cross-check calculations to ensure that results satisfy all given conditions and principles.

Physics problem-solving requires patience and practice but becomes more intuitive with regular effort and understanding of underlying concepts. Always remember to keep track of units and align with conservation laws during calculations.