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Chapter 8: Problem 30

A ball of mass \(m\) sits at the coordinate origin when it explodes into two pieces that shoot along the \(x\) -axis in opposite directions. When one of the pieces (which has mass \(0.270 \mathrm{~m}\) ) is at \(x=70 \mathrm{~cm}\), where is the other piece? [Hint: What happens to the mass center?]

Short Answer

Expert verified
The other piece is approximately at \(-25.9 \, cm\).

Step by step solution

01

Understand the Concept of Center of Mass

Before the explosion, the center of mass of the entire system (ball) is at the origin, because the ball is at rest at the origin. The explosion does not apply any external force, so the center of mass of the system remains at the origin.
02

Set up the Center of Mass Equation

The equation for the center of mass of a system of two particles along the x-axis is given by \(x_{cm} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2}\). In this case, the initial center of mass \(x_{cm0}\) is 0 because the entire ball is at the origin.
03

Apply the Equation to the Final Condition

After the explosion, assume the masses of the pieces are \(m_1 = 0.270m\) at \(x_1 = 0.7 \ m\) (since 70 cm is 0.7 m) and \(m_2 = 0.730m\) at \(x_2\). Set the center of mass equation to equal zero because it must remain at the origin: \(0 = \frac{0.270m \cdot 0.7 + 0.730m \cdot x_2}{0.270m + 0.730m}\).
04

Solve for the Unknown Position

Since the mass \(m\) is present in both terms in the numerator and denominator, it cancels out. So simplify the equation: \[0 = 0.270 \times 0.7 + 0.730 \times x_2\]Now we solve for \(x_2\): \[0.730x_2 = -0.270 \times 0.7\]Calculate the value: \[x_2 = -\frac{0.270 \times 0.7}{0.730}\]Performing the calculations, \[x_2 \approx -0.259 \, m\].
05

Interpret the Result

The negative sign of \(x_2\) indicates that the second piece is located on the negative x-axis, corresponding to the opposite direction to the first piece. Therefore, the position of the second piece is approximately \(-0.259 \, m\), or \(-25.9 \, cm\).

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Key Concepts

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

Explosion Physics
When a ball at rest suddenly explodes into two pieces, a fascinating display of physics known as explosion physics is at play. During an explosion, energy stored within the object is rapidly released, causing it to break apart. Although this seems dramatic, an important principle holds true: the total momentum of the system is conserved.
  • The pieces move in opposite directions because the forces acting during the explosion are internal, meaning they do not alter the total momentum of the system.
  • This results in one piece of the ball having a positive velocity while the other takes an equal but negative velocity, balancing the overall momentum.
This ties into the principle of conservation of momentum. Even in an explosive event, without external forces, the overall motion of a system remains unchanged.
Coordinate System
Understanding the movement of objects after an explosion requires a well-defined coordinate system. When the ball starts at the origin of the axis, the coordinate system assigns the point (0, 0).

In our scenario, we consider a one-dimensional line along the x-axis, simplifying the problem to motion in a straight path. The coordinates tell us the exact location of the ball fragments after the explosion.
  • Before the explosion, both mass and position are straightforward since both are at the origin.
  • After exploding, each piece of the ball adopts a new coordinate based on its direction of movement and speed.
When one piece moves to a position at 0.7 meters along the positive x-axis, it indicates displacement from the origin. The other piece, conversely, moves in the negative direction. Using this straightforward arithmetic within the established coordinate grid helps pinpoint the pieces accurately.
Conservation of Momentum
One of the core principles in this scenario is the conservation of momentum. This principle states that the total momentum of a closed system remains constant if no external forces are applied. In our case, even though the ball explodes, the combined motion of the resulting pieces still adheres to this rule.
  • The center of mass of the system, which initially sat at the origin, must remain there post-explosion.
  • The equation for the center of mass is set as zero (\( x_{cm} = 0 \)) to reflect this consistency.
To solve the problem, we calculated the position of the second piece using the conservation of momentum principle. We equate the initial conditions with the final ones. This involves ensuring that the momentum vectors of the two pieces counterbalance each other, leading to the second fragment sitting at \( x \approx -0.259 \m \).
Thus, the principle of conservation of momentum helps us understand how the system evolves during and after an explosion, maintaining equilibrium overall.

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Most popular questions from this chapter

During a soccer game a ball (of mass \(0.425 \mathrm{~kg}\) ), which is initially at rest, is kicked by one of the players. The ball moves off at a speed of \(26 \mathrm{~m} / \mathrm{s}\). Given that the impact lasted for \(8.0 \mathrm{~ms}\), what was the average force exerted on the ball?

A ball is dropped from a height \(h\) above a tile floor and rebounds to a height of \(0.65 h\). Find the coefficient of restitution between ball and floor. Assign floor quantities the subscript 1 , and ball quantities the subscript 2 . The initial and final velocities of the floor, \(u_{1}\) and \(v_{1}\), are zero. Therefore, $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}=-\frac{v_{2}}{u_{2}} $$ Since we know both the drop and rebound heights \((h\) and \(0.65 h)\), we can write equations for the interchange of \(\mathrm{PE}_{\mathrm{G}}\) and \(\mathrm{KE}\) before and after the impact $$ m g h=\frac{1}{2} m u_{2}^{2} \quad \text { and } \quad m g(0.65 h)=\frac{1}{2} m v_{2}^{2} $$ Therefore, taking down as positive, \(u_{2}=\sqrt{2 g h}\) and \(v_{2}=-\sqrt{1.30 g h}\). Substitution leads to $$ e=\frac{\sqrt{1.30 g h}}{\sqrt{2 g h}}=\sqrt{0.65}=0.81 $$ Notice that the coefficient of restitution equals the square root of the final rebound height over the initial drop height.

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?

An \(8.0\) -g bullet is fired horizontally into a \(9.00\) -kg cube of wood, which is at rest on a frictionless air table. The bullet lodges in the wood. The cube is free to move and has a speed of \(40 \mathrm{~cm} / \mathrm{s}\) after impact. Find the initial velocity of the bullet. This is an example of a completely inelastic collision for which momentum is conserved, although \(\mathrm{KE}\) is not. Consider the system (cube \(+\) bullet). The velocity, and hence the momentum, of the cube before impact is zero. Take the bullet's initial motion to be positive in the positive \(x\) -direction. The momentum conservation law tells us that Momentum of system before impact = Momentum of system after impact $$ \begin{aligned} \text { (Momentum of bullet) }+\text { (Momentum of cube) } &=(\text { Momentum of bullet }+\text { Cube }) \\ m_{B} v_{B x}+m_{C} v_{C x} &=\left(m_{B}+m_{C}\right) v_{x} \\ (0.0080 \mathrm{~kg}) v_{B x}+0 &=(9.008 \mathrm{~kg})(0.40 \mathrm{~m} / \mathrm{s}) \end{aligned} $$ Solving yields \(v_{B x}=0.45 \mathrm{~km} / \mathrm{s}\) and so \(\overrightarrow{\mathbf{v}}_{B}=0.45 \mathrm{~km} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION.

Two balls of equal mass approach the coordinate origin, one moving downward along the \(y\) -axis at \(2.00 \mathrm{~m} / \mathrm{s}\) and the other moving to the right along the \(-x\) -axis at \(3.00 \mathrm{~m} / \mathrm{s}\). After they collide, one ball moves out to the right along the \(+x\) -axis at \(1.20 \mathrm{~m} / \mathrm{s}\). Find the scalar \(x\) and \(y\) velocity components of the other ball. This is a two-dimensional collision and momentum must be conserved independently in each perpendicular direction, \(x\) and \(y\). Take \(u p\) and to the right as positive. Accordingly, keeping in mind that before impact only one ball had an \(x\) -component of velocity, Or $$ \begin{aligned} \text { (Momentum before) }_{x} &=\text { (Momentum after) }_{x} \\ m(3.00 \mathrm{~m} / \mathrm{s})+0 &=m(1.20 \mathrm{~m} / \mathrm{s})+m v_{x} \end{aligned} $$ Here \(v_{\mathrm{x}}\) is the unknown \(x\) -component of velocity of the second ball acquired on impact. Since we know that the first ball lost some of its \(x\) -momentum, the second ball must have gained it. Moreover, (Momentum before) \(_{y}=(\text { Momentum after })_{y}\) or $$ 0+m(-2.00 \mathrm{~m} / \mathrm{s})=0+m v_{y} $$ Here \(v_{\mathrm{y}}\) is the \(y\) -component of velocity of the second ball. (Why the minus sign?) Solving each equation, after cancelling the mass we find that \(v_{x}=1.80 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=-2.00 \mathrm{~m} / \mathrm{s}\)
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