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Chapter 7: Problem 62

A billiard ball moving at a speed of \(2.2 \mathrm{~m} / \mathrm{s}\) strikes an identical stationary ball with a glancing blow. After the collision, one ball is found to be moving at a speed of \(1.1 \mathrm{~m} / \mathrm{s}\) in a direction making a \(60^{\circ}\) angle with the original line of motion. Find the velocity of the other ball.

Short Answer

Expert verified
The velocity of the other ball is approximately 1.9 m/s.

Step by step solution

01

- Understand the problem

A billiard ball strikes another identical ball. After the collision, the moving ball is deflected, and we need to determine the velocity of the second ball post-collision.
02

- Apply the conservation of momentum

Since the system is closed and no external forces are acting, momentum before and after the collision is conserved.
03

- Use momentum components

Set up the conservation of momentum equations in the X and Y directions. For the X direction: \(m v_{1i} = m v_{1f} \cos(60^\circ) + m v_{2f_x}\) and for the Y direction: \(0 = m v_{1f} \sin(60^\circ) - m v_{2f_y}\), where \(v_{1i}\) is the initial velocity of the moving ball, \(v_{1f}\) is the final velocity of the moving ball, and \(v_{2f_x}\) and \(v_{2f_y}\) are the X and Y components of the second ball's velocity.
04

- Solve for \(v_{2f_x}\)

Using the equation for X direction: \(2.2 = 1.1 \cos(60^\circ) + v_{2f_x}\) \(v_{2f_x} = 2.2 - 1.1 \times 0.5) \) \(v_{2f_x} = 1.65\)
05

- Solve for \(v_{2f_y}\)

Using the equation for Y direction: \(0 = 1.1 \sin(60^\circ) - v_{2f_y}\) \(v_{2f_y} = 1.1 \times \frac{\sqrt{3}}{2}\)
06

- Combine the components

Combine the X and Y components to get the velocity vector of the second ball: \(v_{2f} = \sqrt{(v_{2f_x})^2 + (v_{2f_y})^2}\) \(v_{2f} = \sqrt{(1.65)^2 + (0.955)^2}\) \(v_{2f} \approx 1.9\,\mathrm{m/s}\)

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Key Concepts

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

collision analysis
In physics, understanding collisions is crucial as they illustrate the fundamental principles of momentum conservation and energy transfer. When two billiard balls collide, their interaction is dictated by these principles. By examining the pre- and post-event velocities along with angles, we can precisely determine each ball's new trajectory and speed after impact. Initial conditions like speeds and directions critically inform our computations. By leveraging symmetry and identical characteristics of the colliding balls, we can simplify the analysis significantly. In this case, the identical billiard balls allow us to use straightforward momentum equations and trust that the momentum will distribute evenly based on the collision angle and initial speed provided.
momentum components
Breaking down momentum into its components is an essential step in collision analysis, particularly in two-dimensional problems. Here, we decompose the momentum into the X (horizontal) and Y (vertical) directions.
Using the conservation of momentum, we derive two separate equations for these components. For the X direction:
  • \(mv_{1i} = mv_{1f} \, \text{cos}(60^\text{掳}) + mv_{2f_x}\)
And for the Y direction:
  • \(0 = mv_{1f} \, \text{sin}(60^\text{掳}) - mv_{2f_y}\)
Here, the terms \(v_{1i}\) and \(v_{1f}\) represent the initial and final velocities of the first ball, while \(v_{2f_x}\) and \(v_{2f_y}\) represent the X and Y components of the second ball's velocity post-collision. These equations set the stage for solving the unknowns. The solutions can be plugged back into vector calculations to determine the explicit velocity of the second ball after the collision.
velocity vectors
Once we have the momentum components, combining them to find the final velocity is the next step. Velocity vectors give a complete picture of the ball's direction and speed post-collision. To find the resultant velocity, use the Pythagorean theorem since it allows us to integrate the X and Y components effectively.
We compute it as follows: \[v_{2f} = \, \text{sqrt}((v_{2f_x})^2 + (v_{2f_y})^2)\] In our given problem, substituting the calculated components, we find: \[v_{2f} = \, \text{sqrt}((1.65)^2 + (0.955)^2) \,\ \text {\approx 1.9 \ m/s}\] This resultant velocity vector represents the speed and direction of the second ball post-collision. Understanding and determining these vectors are crucial for precisely predicting post-collision trajectories in real-world scenarios like billiards or even vehicle crash investigations.

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Most popular questions from this chapter

Two cars \(A\) and \(B\) slide on an icy road as they attempt to stop at a traffic light. The mass of \(A\) is \(1100 \mathrm{~kg}\), and the mass of \(B\) is \(1400 \mathrm{~kg}\). The coefficient of kinetic friction between the locked wheels of either car and the road is 0.13. Car \(A\) succeeds in stopping at the light, but car \(B\) cannot stop and rear-ends car \(A\). After the collision, \(A\) stops \(8.2 \mathrm{~m}\) ahead of its position at impact, and \(B 6.1 \mathrm{~m}\) ahead; see Fig. 7-27. Both drivers had their brakes locked throughout the incident. Using the material in Chapters 2 and 6 , find the speed of (a) car \(A\) and (b) car \(B\) immediately after impact. (c) Use conservation of translational momentum to find the speed at which car \(B\) struck car \(A\). On what grounds can the use of momentum conservation be criticized here?

Two \(2.0 \mathrm{~kg}\) bodies, \(A\) and \(B\), collide. The velocities before the collision are \(\vec{v}_{A 1}=(15 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(30 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and \(\vec{v}_{B 1}=(-10 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). After the collision, \(\vec{v}_{A 2}=\) \((-5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(20 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). What is the final velocity of \(B\) ?

An object is tracked by a radar station and found to have a position vector given by \(\vec{r}=[(3500 \mathrm{~m})-(160 \mathrm{~m} / \mathrm{s}) t] \hat{\mathrm{i}}+\) \((2700 \mathrm{~m}) \hat{\mathrm{j}}\) with \(\vec{r}\) in meters and \(t\) in seconds. The radar station's \(x\) axis points east, its \(y\) axis north, and its \(z\) axis vertically up. If the object is a \(250 \mathrm{~kg}\) meteorological missile, what are (a) its translational momentum and (b) its direction of motion?

In February 1955, a paratrooper fell \(370 \mathrm{~m}\) from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was \(56 \mathrm{~m} / \mathrm{s}\) (terminal speed), that his mass (including gear) was \(85 \mathrm{~kg}\), and that the magnitude of the force on him from the snow was at the survivable limit of \(1.2 \times 10^{5} \mathrm{~N}\). What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

A rocket sled with a mass of \(2900 \mathrm{~kg}\) moves at \(250 \mathrm{~m} / \mathrm{s}\) on a set of rails. At a certain point, a scoop on the sled dips into a trough of water located between the tracks and scoops water into an empty tank on the sled. By applying the principle of conservation of translational momentum, determine the speed of the sled after \(920 \mathrm{~kg}\) of water has been scooped up. Ignore any retarding force on the scoop.
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