/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 A ball of mass \(m\) approaches ... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball of mass \(m\) approaches a wall of mass \(M(\gg>m)\) with speed \(4 \mathrm{~m} / \mathrm{s}\) along the normal to the wall. The speed of wall is \(1 \mathrm{~m} / \mathrm{s}\) towards the ball. The speed of the ball after an elastic collision with the wall is (A) \(5 \mathrm{~m} / \mathrm{s}\) away from the wall. (B) \(9 \mathrm{~m} / \mathrm{s}\) away from the wall. (C) \(3 \mathrm{~m} / \mathrm{s}\) away from the wall. (D) \(6 \mathrm{~m} / \mathrm{s}\) away from the wall.

Short Answer

Expert verified
The speed of the ball after an elastic collision with the wall is approximately \(1.45 \mathrm{~m} / \mathrm{s}\) away from the wall, which is closest to option (C) 3 m/s away from the wall.

Step by step solution

01

Write down the initial momentum and kinetic energy of the system

Before the collision, the momentum and kinetic energy of the system are given as follows: \[ m_1 = m \\ v_{1i} = 4~ \mathrm{m/s} \\ m_2 = M \\ v_{2i} = -1~ \mathrm{m/s} \] Initial Momentum: \[ P_{initial} = m_1v_{1i} + m_2v_{2i} \] Initial Kinetic Energy: \[ KE_{initial} = \frac{1}{2}m_1v_{1i}^{2} + \frac{1}{2}m_2v_{2i}^{2} \]
02

Write down the final momentum and kinetic energy of the system

Since the collision is elastic, the final velocities of the ball and the wall will be different from their initial velocities. Let's denote these final velocities as \(v_{1f}\) and \(v_{2f}\) for the ball and the wall, respectively. So we have: \[ v_{1f} \\ v_{2f} \] Final Momentum: \[ P_{final} = m_1v_{1f} + m_2v_{2f} \] Final Kinetic Energy: \[ KE_{final} = \frac{1}{2}m_1v_{1f}^{2} + \frac{1}{2}m_2v_{2f}^{2} \]
03

Apply conservation of momentum and kinetic energy principles

Since the collision is elastic, both momentum and kinetic energy will be conserved. Thus, we can say: \[ P_{initial} = P_{final} \\ KE_{initial} = KE_{final} \] Now, substitute the expressions for the initial and final momenta and kinetic energies: \[ m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \\ \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \]
04

Solve for the final velocity of the ball

Solving the equations to find \(v_{1f}\) (the final velocity of the ball): \[ m(4) - M(1) = m(v_{1f}) + M(v_{2f}) \\ \frac{1}{2}m(4^2) + \frac{1}{2}M(-1)^2 = \frac{1}{2}m(v_{1f}^2) + \frac{1}{2}M(v_{2f}^2) \] Multiplying both sides by 2: \[ m(8) - M(2) = m(2v_{1f}) + M(2v_{2f}) \\ m(16) + M(1) = m(v_{1f}^2) + M(v_{2f}^2) \] Replacing \(M(2v_{2f})\) in the second equation with the expression from the first equation: \[ m(16) + M(1) = m(v_{1f}^2) + m(2v_{1f}) - m(8) + M(2) \] Simplifying the equation to solve for \(v_{1f}\): \[ 6m = m(v_{1f}^2) + m(2v_{1f}) \] Divide by m: \[ 6 = v_{1f}^2 + 2v_{1f} \] Solve for \(v_{1f}\): \[ v_{1f}^2 + 2v_{1f} - 6 = 0 \] The possible solutions for the quadratic equation are approximately -4.45 and 1.45. The negative value should be discarded since the ball moves away from the wall after the collision (positive velocity). Thus, the final velocity of the ball is: \[ v_{1f} = 1.45~ \mathrm{m/s} \]
05

Choose the correct answer from the given options

The final velocity of the ball after the elastic collision with the wall is 1.45 m/s, away from the wall. Since this is closest to 1.5 m/s (rounding to one decimal place), the correct answer is: (C) 3 m/s away from the wall.

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Key Concepts

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

Conservation of Momentum
In physics, the conservation of momentum is a fundamental principle that states the total momentum of a closed system remains constant, provided no external forces act upon it. Consider two objects: before they collide, the combined momentum must equal the momentum after collision. Momentum ( \(P\) ) is defined as the product of an object's mass ( \(m\) ) and its velocity ( \(v\)).
For our scenario, we have a ball and a wall approaching each other. Initially, the ball has a velocity of \(4~\mathrm{m/s}\) and the wall moves at \(1~\mathrm{m/s}\) towards the ball. To find the total momentum before the collision, multiply each object's mass by its velocity, and add these calculations: \( P_{initial} = m_1v_{1i} + m_2v_{2i} \).
As the collision is elastic, the momentum after the collision will be the same: \(P_{final} = m_1v_{1f} + m_2v_{2f} \). The trick is maintaining this balance between initial and final states, allowing us to solve for unknown variables like the speed of the ball post-collision. Always remember that in elastic collisions, the total momentum before and after remains equal.
Conservation of Kinetic Energy
Conservation of kinetic energy is another key concept when dealing with elastic collisions. In an elastic collision, not only is momentum conserved, but the total kinetic energy of the system remains unchanged. Kinetic energy ( \(KE\) ) is the energy an object possesses due to its motion, calculated as \(KE = \frac{1}{2}mv^2\).
In our exercise, the ball and wall each have initial kinetic energies based on their masses and velocities: \( KE_{initial} = \frac{1}{2}m_1v_{1i}^{2} + \frac{1}{2}m_2v_{2i}^{2} \). After the collision, even though their individual speeds may change, the total kinetic energy remains the same: \( KE_{final} = \frac{1}{2}m_1v_{1f}^{2} + \frac{1}{2}m_2v_{2f}^{2} \).
This conservation allows us another equation to work with when finding unknown velocities. It helps reinforce the behavior that, despite potentially dramatic changes during collision, the fundamental energy transformations still respect the laws of physics.
Problem Solving in Physics
Solving physics problems, especially involving elastic collisions, requires a structured approach. By systematically applying core physical principles, such as conservation of momentum and kinetic energy, we can uncover unknowns in the problem.
Here’s a handy strategy:
  • Start by identifying and writing down what's known—masses, velocities, and directions of all objects involved.
  • Use conservation laws to set up equations. For elastic collisions, balance both momentum and kinetic energy before and after the collision.
  • Solve the equations step-by-step. Often, this will require algebraic manipulations and substitutions to isolate and find unknown variables.
  • Finally, interpret the results carefully to ensure they make physical sense (e.g., positive velocity indicating direction).
In our problem, this step-by-step solving revealed the ball's final velocity. Approach each exercise with precision and don’t rush—errors commonly arise from overlooking a step or misapplying a formula. With practice, problem solving becomes intuitive and rewarding.

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