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Chapter 8: Problem 13

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 up and to the right as positive. Accordingly, keeping in mind that before impact only one ball had an \(x\) -component of velocity, or \(m(3,00 \mathrm{~m} / \mathrm{s})+0=m(1.20 \mathrm{~m} / \mathrm{s}+m v\) Here \(u_{x}\) 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, \(0 r\) Here \(u_{y}\) 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 \(v_{y}=-2.00 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The second ball has velocities: \( u_{2x} = -1.80 \ m/s \) and \( u_{2y} = -2.00 \ m/s \).

Step by step solution

01

Identify Initial and Final Momentum

Initially, we have two balls each of mass \( m \). For the first ball, moving down along the \(y\)-axis, its initial velocity \( v_{1y} = -2.00 \ m/s \) and \( u_{1x} = 0 \ m/s \). For the second ball, moving along the \(-x\)-axis, its initial velocity \( v_{2x} = -3.00 \ m/s \) and \( u_{2y} = 0 \ m/s \).After the collision, the first ball has \( u_{1x} = 1.20 \ m/s \) (since it moves to the right) and the unknown \( u_{1y} \). The second ball has the unknown final velocities \( u_{2x} \) and \( u_{2y} \).
02

Apply Conservation of Momentum in x-direction

Using the conservation of momentum in the \(x\) direction:\[ m(-3.00) + m(0) = m(1.20) + m(u_{2x}) \]Since their masses are equal, we can cancel \( m \) from all terms:\[ -3.00 = 1.20 + u_{2x} \]Solving for \( u_{2x} \), we find:\[ u_{2x} = -3.00 - 1.20 = -1.80 \ m/s \]
03

Apply Conservation of Momentum in y-direction

Using the conservation of momentum in the \(y\) direction:\[ m(0) + m(-2.00) = m(0) + m(u_{2y}) \]Again, we can cancel \( m \) from all terms:\[ -2.00 = u_{2y} \]Thus, the \( y \)-component of the velocity of the second ball is \( u_{2y} = -2.00 \ m/s \).
04

Interpret the Results

The calculations show that the second ball moves with a velocity component of \(-1.80 \ m/s\) along the \( x \)-axis and \(-2.00 \ m/s\) along the \( y \)-axis. The negative signs indicate that the second ball moves left and downward.

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Key Concepts

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

Two-dimensional collision
When discussing two-dimensional collisions, you're looking at encounters between objects moving in a 2D plane. This is a bit more complex than dealing with one-dimensional collisions, where objects tend to move along a single line. In a two-dimensional collision, each object can possess velocities and momenta in both the x and y directions.
This means you need to analyze movements based on these perpendicular directions as they鈥檙e independent of each other.
  • Imagine two balls moving toward each other across a flat surface at given angles. They collide, bounce off, and continue in new directions.
  • Your task is to determine where they go and the speed of each after the collision.
To solve problems involving two-dimensional collisions, you consider not just the speeds but importantly, the directions they travel. Conservation of momentum is a fundamental principle in these scenarios, meaning the total momentum before the collision equals the total momentum after, even when looked at separately in x and y components.
Velocity components
Velocity components break down the movement of objects into simpler, single-directional parts. In the given exercise, each ball鈥檚 movement is reduced to components along the x and y axes.
In physics, velocity is described both by speed and direction. When represented graphically or in mathematical terms, you use vectors. Each vector can be split into two perpendicular components.
  • The ball moving downward along the y-axis at 2.00 m/s has x-component 0 and y-component -2.00 m/s (negative since it's moving down).
  • The one moving left along the x-axis at 3.00 m/s has y-component 0 and x-component -3.00 m/s (negative since it鈥檚 to the left).
This simplification allows you to apply formulas much more easily, focusing on each direction independently.
Momentum conservation equations
The momentum conservation equations are central to solving collision problems. They help assert that total linear momentum in each direction is conserved through the collision process.
Breaking it down further, you consider the momentum equation separately in the x and y directions, as seen in the given problem.
  • For the x-direction: Set up the equation based on initial and final momenta with respect to only x components. Here, you'll cancel the equal masses and solve for the unknown final velocity components.
  • For the y-direction: Similarly, use the initial y-momentum equalling the final y-momentum to compute unknowns.
These equations reinforce that the physics is true regardless of the complexities involving angles and directions after impact.
Collision problem-solving steps
Solving any collision problem methodically is key to finding correct solutions. Let鈥檚 walk through the steps:
  • Identify all initial conditions with clear definitions of velocity components for each object involved.
  • Apply conservation laws to set up post-collision equations. Do this separately for momentum in x and y directions.
  • Solve the equations progressively to find unknown velocity components.
  • Analyze the results: The signs of the velocity components give insights into the directions of movement post-collision. A negative sign can indicate a reversed direction, such as moving left or down.
These steps allow you a clear pathway to assess and solve two-dimensional collision problems systematically and with confidence.

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Most popular questions from this chapter

A billiard ball moving at a speed \(u_{1 i}\) strikes, head-on, another billiard ball that is at rest. Assuming the collision is completely elastic, show that $$ v_{1 i}^{2}=v_{\mathrm{lf}}^{2}+v_{2 f}^{2} $$

Three point masses are placed on the \(x\) -axis: \(200 \mathrm{~g}\) at \(x=0,500 \mathrm{~g}\) at \(x=30 \mathrm{~cm}\), and \(400 \mathrm{~g}\) at \(x=70 \mathrm{~cm}\). Find their center of mass. We can make the calculation with respect to any point, but since all the data is measured from the \(x=0\) origin, that point will do nicely. $$ x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(0.20 \mathrm{~kg})+(0.30 \mathrm{~m})(0.50 \mathrm{~kg})+(0.70 \mathrm{~m})(0.40 \mathrm{~kg})}{(0.20+0.50+0.40) \mathrm{kg}}=0.39 \mathrm{~m} $$ The center of mass is located at a distance of \(0.39 \mathrm{~m}\), in the positive \(x\) -direction, from the origin. The \(y\) - and z-coordinates of the center of mass are zero.

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 slow 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 from which $$ \begin{array}{c} F \Delta t=m v_{f}-m v_{i}=m(50000 \mathrm{~m} / \mathrm{s})-0 \\ F=(50000 \mathrm{~m} / \mathrm{s}) \frac{\mathrm{m}}{\mathrm{L}} \end{array} $$ 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 $$ F=(50000 \mathrm{~m} / \mathrm{s})(1500 \mathrm{~kg} / \mathrm{s})=75 \mathrm{MN} $$ 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 $$ M_{\text {rocket }}=\frac{\text { weight }}{g}=\frac{75 \times 10^{6} \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=7.7 \times 10^{6} \mathrm{~kg} $$

What force is exerted on a stationary flat plate held perpendicular to a jet of water as shown in Fig. 8-6? The horizontal speed of the water is \(80 \mathrm{~cm} / \mathrm{s}\), and \(30 \mathrm{~mL}\) of the water hit the plate each second. Assume the water moves parallel to the plate after striking it. One milliliter (mL) of water has a mass of \(1.00 \mathrm{~g}\). This question deals with speed, mass, time, and force, and that suggests impulse-momentum and Newton's Second Law. The plate exerts an impulse on the water and changes its horizontal momentum. The water exerts a counterforce on the plate. Taking the direction to the right as positive, (Impulse) \(_{x}=\) Change in \(x\) -directed momentum $$ F_{x} \Delta t=\left(m v_{x}\right)_{\text {final }}-\left(m v_{x}\right)_{\text {initial }} $$ Let \(t\) be \(1.00 \mathrm{~s}\) so that \(m\) will be the mass that strikes in \(1.00 \mathrm{~s}\), namely \(30 \mathrm{~g}\). Then the above equation becomes $$ F_{x}(1.00 \mathrm{~s})=(0.030 \mathrm{~kg})(0 \mathrm{~m} / \mathrm{s})-(0.030 \mathrm{~kg})(0.80 \mathrm{~m} / \mathrm{s}) $$ from which \(F_{x}=-0.024 \mathrm{~N}\). This is the force exerted by the plate on the water. The law of action and reaction tells us that the jet exerts an equal but opposite force on the plate.

A 16 -g mass is moving in the \(+x\) -direction at \(30 \mathrm{~cm} / \mathrm{s}\), while a \(4.0\) -g mass is moving in the \(-x\) -direction at \(50 \mathrm{~cm} / \mathrm{s}\). They collide headon and stick together. Find their velocity after the collision. Assume negligible friction. This is a completely inelastic collision for which KE is not conserved, although momentum is. Let the 16 -g mass be \(m_{1}\) and the 4.0-g mass be \(m_{2}\). Take the \(+x\) -direction to be positive. That means that the velocity of the \(4.0\) -g mass has a scalar value of \(u_{2 x}\) \(=-50 \mathrm{~cm} / \mathrm{s}\). We apply the law of conservation of momentum to the system consisting of the two masses: $$ \begin{array}{l} \text { Momentum before impact = Momentum after impact }\\\ \begin{aligned} m_{1} v_{1 x}+m_{2} v_{2 x} &=\left(m_{1}+m_{2}\right) v_{x} \\ (0.016 \mathrm{~kg})(0.30 \mathrm{~m} / \mathrm{s})+(0.0040 \mathrm{~kg})(-0.50 \mathrm{~m} / \mathrm{s}) &=(0.020 \mathrm{~kg}) v_{x} \\ v_{x} &=+0.14 \mathrm{~m} / \mathrm{s} \end{aligned} \end{array} $$ (Notice that the \(4.0\) -g mass has negative momentum.) Hence, \(\overrightarrow{\mathbf{v}}=\) \(0.14 \mathrm{~m} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION
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