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The coefficient of restitution is 0.9 between the two 60 -mm-diameter billiard balls \(A\) and \(B .\) Ball \(A\) is moving in the direction shown with a velocity of \(1 \mathrm{m} / \mathrm{s}\) when it strikes ball \(B\), which is at rest. Knowing that after impact \(B\) is moving in the \(x\) direction, determine (a) the angle \(\theta,(b)\) the velocity of \(B\) after impact.

Short Answer

Expert verified
\(\theta = 0\), \(v_B = 0.9 \text{ m/s}\).

Step by step solution

01

Define the Problem

We need to find the angle \(\theta\) and the velocity of billiard ball \(B\) after impact, given the initial conditions and the coefficient of restitution is 0.9.
02

Apply the Coefficient of Restitution Formula

The coefficient of restitution \(e\) is given by the equation \(e = \frac{v_B - v_A}{u_A - u_B}\), where \(v_B\) and \(v_A\) are the velocities of balls \(B\) and \(A\) after collision, and \(u_A = 1 \text{ m/s}\) and \(u_B = 0\) are the initial velocities. Rearrange the formula to express the relationship between the velocities.
03

Conserve Momentum in the x-Direction

Using conservation of momentum in the x-direction, we have \(m_A \cdot u_A = m_A \cdot v_A \cdot \cos\theta + m_B \cdot v_B\). Note that the masses \(m_A\) and \(m_B\) are equal, thus they cancel out. Simplifying gives us the equation for the x-direction: \(1 = v_A \cdot \cos\theta + v_B\).
04

Conserve Momentum in the y-Direction

In the y-direction, the initial momentum is zero, equating to the sum of the components of the velocities after impact: \(0 = m_A \cdot v_A \cdot \sin\theta\). Therefore, \(v_A \cdot \sin\theta = 0\), leading to \(v_A = 0\) or \(\sin\theta = 0\), indicating no momentum in the y-direction of \(B\).
05

Solve the System of Equations

Substitute the restitution equation and momentum equations together: \[ e = 0.9 = \frac{v_B}{1} \] to give \(v_B = 0.9\). Substitute \(v_B\) into the x-direction momentum equation: \[ 1 = v_A \cdot \cos\theta + 0.9 \]. Since \(v_A = 0.1\), this implies \(\cos\theta \approx 1\), resulting \(\theta = 0\) or \(v_A \approx 0.1\, \text{m/s}.\)
06

Conclusions

The results indicate that ball \(B\) moves solely in the x-direction after the impact, while ball \(A\) continues to move with minimal velocity at an angle of 0 degrees to the x-axis.

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Key Concepts

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

Conservation of Momentum
Momentum is a fundamental concept in physics, especially when analyzing collisions. The principle of conservation of momentum states that if no external forces act on a system, the total momentum of the system remains constant before and after a collision.

In the exercise involving billiard balls, the conservation of momentum helps us understand the behavior of these balls post-impact. To apply this concept:
  • Consider the momentum in both the x and y directions separately.
  • In this example, only the x-direction has initial momentum due to Ball A moving with a velocity of 1 m/s.
  • The y-direction momentum begins at zero since Ball B is stationary and there's no motion in that direction initially.
By setting up equations for each direction and knowing the mass is equal for both balls, we can cancel the mass terms and simplify our calculations.

This gives us insight into how energy and motion transfer between the balls, making the solution to the problem more approachable.
Collision Analysis
Collision analysis involves understanding how objects behave when they crash into each other. There are two main types of collisions: elastic and inelastic.

In this exercise, we deal with nearly elastic collisions indicated by the coefficient of restitution of 0.9. This number shows that most, but not all, kinetic energy is conserved during the impact.

Key steps in analyzing this situation include:
  • Determining the post-impact velocities using the restitution coefficient.
  • Calculating the angles of motion after the collision.
  • Ensuring momentum is conserved as per physical laws.
In our scenario, we applied the given coefficient to derive Ball B's velocity. With momentum conserved in both directions, a system of equations allows solving for unknowns like angle and velocity.

Understanding this process is essential for solving collision-related problems in physics, whether dealing with real-world scenarios or theoretical models.
Impact Velocity
The impact velocity refers to the speed at which two objects come into contact during a collision. In the exercise, Ball A strikes Ball B with a set velocity, catalyzing the analysis of their movements post-collision.

This concept requires knowing the initial velocities of colliding bodies to predict outcomes:
  • Initial velocity of Ball A is 1 m/s, while Ball B is at rest with 0 m/s.
  • Impact velocity is crucial for determining resultant velocities using the restitution coefficient.
  • Affects how momentum transfers between the balls, affecting both velocity and direction post-collision.
In practice, impact velocity influences many real-world outcomes, such as vehicle crashes or sports mechanics. Therefore, by mastering these calculations, one gains insights into how different materials and angles affect the results of collisions, preparing them not only for exercises but practical applications as well.

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Most popular questions from this chapter

Block A is released from rest and slides down the frictionless surface of B until it hits a bumper on the right end of B. Block A has a mass of 10 kg and object B has a mass of 30 kg and B can roll freely on the ground. Determine the velocities of A and B immediately after impact when (a) e 5 0, (b) e 5 0.7.

A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a constant force of 2500 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5 m/s.

Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop. A 10 -ton truck enters a \(15^{\circ}\) ramp at a high speed \(v_{0}=108 \mathrm{ft} / \mathrm{s}\) and travels for \(6 \mathrm{s}\) before its speed is reduced to \(36 \mathrm{ft} / \mathrm{s}\). Assuming constant deceleration, determine \((a)\) the magnitude of the braking force, \((b)\) the additional time required for the truck to stop. Neglect air resistance and rolling resistance.

A 5 -kg sphere is dropped from a height of \(y=2 \mathrm{m}\) to test newly designed spring floors used in gymnastics. The mass of the floor section is \(10 \mathrm{kg}\), and the effective stiffness of the floor is \(k=120 \mathrm{kN} / \mathrm{m}\). Knowing that the coefficient of restitution between the ball and the platform is \(0.6,\) detrmine \((a)\) the height \(h\) reached by the sphere after rebound, \((b)\) the maximum force in the springs.

At an amusement park there are 200 -kg bumper cars \(A, B,\) and \(C\) that have riders with masses of \(40 \mathrm{kg}, 60 \mathrm{kg},\) and \(35 \mathrm{kg}\), respectively. Car \(A\) is moving to the right with a velocity \(\mathrm{v}_{A}=2 \mathrm{m} / \mathrm{s}\) when it hits stationary car \(B\). The coefficient of restitution between each car is \(0.8 .\) Determine the velocity of car \(C\) so that after car \(B\) collides with car \(C\) the velocity of \(\operatorname{car} B\) is zero.

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