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Chapter 8: Problem 31

A ball of mass \(m\) at rest at the coordinate origin explodes into three equal pieces. At some instant, one piece is on the \(x\) -axis at \(x=40 \mathrm{~cm}\) and another is at \(x=20 \mathrm{~cm}, y=-60 \mathrm{~cm}\). Where is the third piece at that instant?

Short Answer

Expert verified
The third piece is at \((-60 \text{ cm}, 60 \text{ cm})\).

Step by step solution

01

Understand the System and Conservation Law

We have a system of three identical pieces of a ball after it exploded. The total momentum of the system must be conserved. Since the ball was initially at rest, the total initial momentum is zero.
02

Assign Coordinates and Masses

Let's denote the masses of the pieces as \(m_1 = m_2 = m_3 = \frac{m}{3}\), because the pieces are of equal mass. The coordinates for the first piece are \((40 \text{ cm}, 0)\) and for the second piece is \((20 \text{ cm}, -60 \text{ cm})\). We need to find the coordinates \((x_3, y_3)\) of the third piece.
03

Apply Conservation of Momentum along the X-axis

Since the initial momentum along the x-axis is zero, the sum of the momenta of all pieces along the x-axis must also be zero. This gives us: \[ m_1 \cdot 40 + m_2 \cdot 20 + m_3 \cdot x_3 = 0 \]. Simplifying, we use \(m_1 = m_2 = m_3 = \frac{m}{3}\), hence: \[ \frac{m}{3} \cdot 40 + \frac{m}{3} \cdot 20 + \frac{m}{3} \cdot x_3 = 0 \]. Simplifying the equation further: \[ 40 + 20 + x_3 = 0 \], therefore \(x_3 = -60\).
04

Apply Conservation of Momentum along the Y-axis

Using a similar logic for the y-axis and knowing the initial momentum is zero, the equation is: \[ m_1 \cdot 0 + m_2 \cdot (-60) + m_3 \cdot y_3 = 0 \]. Simplifying, \(m_2 = m_3 = \frac{m}{3}\), therefore \[ \frac{m}{3} \cdot (-60) + \frac{m}{3} \cdot y_3 = 0 \]. Simplify to find: \[ -60 + y_3 = 0 \], so \(y_3 = 60\).
05

Final Coordinates of the Third Piece

From the above steps, it is determined that the coordinates of the third piece are \((-60 \text{ cm}, 60 \text{ cm})\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Physics Problem Solving
When tackling physics problems, especially those involving conservation laws, a structured approach is valuable. Start with understanding the system and what principles apply, such as conservation of momentum in this case. Here, the ball initially at rest explodes into three pieces. This scenario suggests that the overall momentum before the explosion is zero, setting the stage for applying momentum conservation principles.
  • Break down the problem: Identify known and unknown variables.
  • List the laws or principles, such as conservation of momentum, that can be applied.
  • Approach the problem step-by-step, applying the principles logically to find the solution.
  • Assign coordinates to all moving parts in the system to solve for unknowns.
With each piece having equal mass and initial total momentum of zero, the individual momenta must balance out so that their sum remains zero. This approach not only helps solve such exercises but also strengthens understanding of fundamental physics concepts.
Momentum Conservation in Explosions
Momentum conservation is pivotal when analyzing explosions. An explosion causes objects originally at rest to fly apart, changing their positions and velocities. Here, the main focus is understanding how conservation of momentum applies.
Since the ball was initially at rest, the total momentum before it exploded was zero. Therefore, the vector sum of the momentum of the pieces post-explosion also must equal zero.
  • Each piece of the ball after the explosion exerts momentum in a specific direction.
  • The conservation of momentum dictates that the momentum in every direction must collectively add up to zero.
For each fragmented piece:- The x-direction and y-direction are considered separately in this problem.- Use equations like \[ m_1 \cdot v_{x1} + m_2 \cdot v_{x2} + m_3 \cdot v_{x3} = 0 \] to determine unknown velocities or positions.Apply such equations carefully to track changes post-explosion and solve for unknown positions or velocities, as demonstrated with the third fragment.
Coordinate Geometry in Physics
Coordinate geometry helps in mapping and solving problems involving objects in motion or at specific positions. Physics exercises like the exploding ball problem rely heavily on coordinate geometry to determine the locations of objects post-event.
  • Assign coordinates to known points, as seen where two pieces land at \((40 \text{ cm}, 0)\) and \((20 \text{ cm}, -60 \text{ cm})\).
  • Calculate unknown coordinates by utilizing conservation principles and equating sums aligned with coordinate directions.
The goal is to use equations derived from initial conditions and mathematical concepts to pinpoint the location of unknowns. Here:- Solving for the x-coordinate using \[ 40 + 20 + x_3 = 0 \] helps locate the third piece on the x-axis.- Similarly, for the y-coordinate, use \[ -60 + y_3 = 0 \] to find its position on the y-axis.These methods illustrate how physics interplays with mathematical tools to achieve solutions.

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Most popular questions from this chapter

What average resisting force must act on a \(3.0-\mathrm{kg}\) mass to reduce its speed from \(65 \mathrm{~cm} / \mathrm{s}\) to \(15 \mathrm{~cm} / \mathrm{s}\) in \(0.20 \mathrm{~s}\) ?

A \(15-\mathrm{g}\) bullet moving at \(300 \mathrm{~m} / \mathrm{s}\) passes through a \(2.0-\mathrm{cm}\) -thick sheet of foam plastic and emerges with a speed of \(90 \mathrm{~m} / \mathrm{s}\). Assuming that the speed change takes place uniformly, what average force impeded the bullet's motion through the plastic? We can determine the change in momentum, and that suggests using the impulse equation to find the force \(F\) on the bullet as it takes a time \(\Delta t\) to pass through the plastic. Taking the initial direction of motion to be positive, $$ F \Delta t=m v_{f}-m v_{i} $$ We can find \(\Delta t\) by assuming uniform deceleration and using \(x=v_{a v} t\), where \(x=0.020 \mathrm{~m}\) and \(v_{a v}=\frac{1}{2}\left(v_{i}+v_{f}\right)=\) \(195 \mathrm{~m} / \mathrm{s}\). This gives \(\Delta t=1.026 \times 10^{-4} \mathrm{~s}\). Then $$ (F)\left(1.026 \times 10^{-4} \mathrm{~s}\right)=(0.015 \mathrm{~kg})(90 \mathrm{~m} / \mathrm{s})-(0.015 \mathrm{~kg})(300 \mathrm{~m} / \mathrm{s}) $$ which yields \(F=-3.1 \times 10^{4} \mathrm{~N}\) as the average retarding force. How could this problem have been solved using \(F=m a\) instead of the impulse equation? By using energy methods?

A \(2.0\) -kg brick is moving at a speed of \(6.0 \mathrm{~m} / \mathrm{s}\). How large a force \(F\) is needed to stop the brick in a time of \(7.0 \times 10^{-4} \mathrm{~s}\) ? Since we have a force and the time over which it acts, that suggests using the impulse equation (i.e., Newton's Second Law): Impulse on brick \(=\) Change in momentum of brick $$ \begin{aligned} F \Delta t &=m v_{f}-m v_{i} \\ F\left(7.0 \times 10^{-4} \mathrm{~s}\right) &=0-(2.0 \mathrm{~kg})(6.0 \mathrm{~m} / \mathrm{s}) \end{aligned} $$ from which \(F=-1.7 \times 10^{4} \mathrm{~N}\). The minus sign indicates that the force opposes the motion.

A rocket standing on its launch platform points straight upward. Its engines are activated and eject gas at a rate of \(1500 \mathrm{~kg} / \mathrm{s}\). The molecules are expelled with an average speed of \(50 \mathrm{~km} / \mathrm{s}\). How much mass can the rocket initially have if it is slowly to rise because of the thrust of the engines? The problem provides mass flow and speed, the product of which is equivalent to the time rate-of-change of momentum. That should bring to mind the impulse- momentum relationship, which, of course, is Newton's Second Law. Since the initial motion of the rocket itself is negligible in comparison to the speed of the expelled gas, we can assume the gas is accelerated from rest to a speed of \(50 \mathrm{~km} / \mathrm{s}\). The impulse required to provide this acceleration to a mass \(m\) of gas is $$ F \Delta t=m v_{f}-m v_{i}=m(50000 \mathrm{~m} / \mathrm{s})-0 $$ from which $$ F=(50000 \mathrm{~m} / \mathrm{s}) \frac{m}{\Delta t} $$ But we are told that the mass ejected per second \((m / \Delta t)\) is \(1500 \mathrm{~kg} / \mathrm{s}\), and so the force exerted on the expelled gas is $$ F=(50000 \mathrm{~m} / \mathrm{s})(1500 \mathrm{~kg} / \mathrm{s})=75 \mathrm{MN} $$ An equal but opposite reaction force acts on the rocket, and this is the upward thrust on the rocket. The engines can therefore support a weight of \(75 \mathrm{MN}\), so the maximum mass the rocket could have is $$ M_{\text {rocket }}=\frac{\text { weight }}{g}=\frac{75 \times 10^{6} \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=7.7 \times 10^{6} \mathrm{~kg} $$

Two balls of equal mass, moving with speeds of \(3 \mathrm{~m} / \mathrm{s}\), collide head-on. Find the speed of each after impact if (a) they stick together, \((b)\) the collision is perfectly elastic, \((c)\) the coefficient of restitution is \(1 / 3\).
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