Question

A shell traveling with speed $v_{\mathrm{o}}$ exactly horizontally and due north explodes into two equal-mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed $v_{\mathrm{o}}.$ What is the velocity of the other fragment?

Short answer

The second fragment's velocity is $2v_{\mathrm{o}}$ horizontally.

Step-by-step solution

Understand the Principle

In this exercise, we will use the principle of conservation of momentum, which states that the total momentum of a system remains constant, as no external forces are acting on it. Initially, the only momentum is due to the shell moving horizontally with velocity $v_{\mathrm{o}}$.

Initial Momentum Analysis

Calculate the initial momentum of the system. Since the shell is moving horizontally with velocity $v_{\mathrm{o}}$ and the total mass is $2m$ (where $m$ is the mass of one fragment), the initial momentum $p_i$ is: $$p_i=(2m)v_{\mathrm{o}}.$$

Momentum Conservation Equation

After the explosion, one fragment moves vertically upwards with velocity $v_{\mathrm{o}}$ and mass $m$. Thus, its momentum is: $$p_1=m\cdot v_{\mathrm{o}}.$$ The conservation of momentum in the horizontal direction gives:$$(2m)v_{\mathrm{o}}=m\cdot v_{\mathrm{x}}+0,$$ where $v_{\mathrm{x}}$ is the horizontal velocity of the other fragment.

Solve for Horizontal Velocity

From the equation:$$(2m)v_{\mathrm{o}}=m\cdot v_{\mathrm{x}},$$solve for $v_{\mathrm{x}}$:$$v_{\mathrm{x}}=2v_{\mathrm{o}}.$$This is the horizontal velocity of the second fragment.

Final Velocity of the Second Fragment

Since no vertical forces act during the explosion, the vertical velocity of the second fragment is 0. Thus, the total velocity $v$ of the second fragment is: $$v=2v_{\mathrm{o}}\hat{i}+0\hat{j},$$or simply, the second fragment travels at $2v_{\mathrm{o}}$ horizontally.

Key concepts

Explosion Dynamics

In physical terms, an explosion can be thought of as a rapid expansion of matter in which the energy is suddenly released. This change causes the original object to break apart, typically influenced by physical laws such as Newton's laws of motion. When a shell explodes, as in the given problem, the explosion distributes the stored energy among the fragments. Because the explosion happens in an isolated system (no external forces act on it), this means two things: first, the momentum before the explosion is equal to the total momentum after the explosion; second, each fragment will receive energy that propels it in different directions, but the energy and momentum before and after will still relate through conservation laws. In this example, the shell initially moves horizontally due north, and when it explodes, its energy causes the creation of two fragments, which move in directions and with velocities governed by these conservation laws. This underscores how energy transfer and momentum conservation play integral roles in explosion dynamics.

Momentum Conservation Equation

The principle of momentum conservation is a fundamental concept in physics. It states that if no external force acts on a system, the system's total momentum remains constant. Momentum, defined as the product of mass and velocity, plays a crucial role in analyzing systems such as exploding shells.In this exercise, initially, the momentum of the whole system (both fragments together) was carried by the shell traveling horizontally with velocity $v_{\mathrm{o}}$. The total initial momentum was $(2m)v_{\mathrm{o}}$, given that the shell's total mass before the explosion was $2m$.Upon explosion, one fragment travels vertically upwards with velocity $v_{\mathrm{o}}$, meaning its momentum is $m{v}_{\mathrm{o}}$. According to the principle of conservation of momentum, the momentum must remain constant in both horizontal and vertical directions separately. Therefore, the equation that describes this conservation in the horizontal direction is simple and follows from the initial problem:$$(2m)v_{\mathrm{o}}=m\cdot v_{\mathrm{x}}+0.$$This states that the initial momentum distributes between both fragments such that the horizontal component of momentum remains unchanged after the explosion.

Horizontal Velocity Calculation

After understanding the conservation of momentum, the task in this exercise was to determine the velocity of the second fragment after the explosion. With one fragment moving vertically, all the initial horizontal momentum must be carried by the second fragment, as it retains all the horizontal velocity present before the explosion.Since the initial horizontal momentum was $(2m)v_{\mathrm{o}}$, and the fragment moving vertically contributes no horizontal component, the entire horizontal momentum must be conserved by the alternate fragment. Using the equation derived from momentum conservation:$$(2m)v_{\mathrm{o}}=m\cdot v_{\mathrm{x}},$$we can solve for the horizontal velocity, $v_{\mathrm{x}}$, of the second fragment. Rearranging gives:$$v_{\mathrm{x}}=\frac{(2m)v_{\mathrm{o}}}{m}=2v_{\mathrm{o}}.$$This calculation shows that the second fragment travels exactly twice the initial horizontal speed of the shell. This result clearly illustrates how momentum conservation dictates the velocity distribution post-explosion, crucial for understanding the behavior of moving and exploding objects in physics.