91Ó°ÊÓ

Two identical balls collide head-on. The initial velocity of one is \(0.75 \mathrm{~m} / \mathrm{s}\) - EAST, while that of the other is \(0.43 \mathrm{~m} / \mathrm{s}\) -WEST. If the collision is perfectly elastic, what is the final velocity of each ball? Because the collision is perfectly elastic, both momentum and KE are conserved. Since the collision is head-on, all motion takes place along a straight line. Take east as positive and call the mass of each ball \(m\). Momentum is conserved in a collision, so we can write Momentum before \(=\) Momentum after $$ m(0.75 \mathrm{~m} / \mathrm{s})+m(-0.43 \mathrm{~m} / \mathrm{s})=m v_{1}+m v_{2} $$ where \(v_{1}\) and \(v_{2}\) are the final values. This equation simplifies to $$ 0.32 \mathrm{~m} / \mathrm{s}=v_{1}+v_{2} $$ Because the collision is assumed to be perfectly elastic, KE is also conserved. Thus, $$ \begin{array}{c} \text { KE before }=\text { KE after } \\ \frac{1}{2} m(0.75 \mathrm{~m} / \mathrm{s})^{2}+\frac{1}{2} m(-0.43 \mathrm{~m} / \mathrm{s})^{2}=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} m v_{2}^{2} \end{array} $$ This equation can be simplified to $$ 0.747=v_{1}^{2}+v_{2}^{2} $$ We can solve for \(u_{2}\) in Eq. (1) to get \(v_{2}=0.32-v_{1}\) and substitute that into Eq. (2). This yields $$ 0.747=\left(0.32-v_{1}\right)^{2}+v_{1}^{2} $$ \(2 v_{1}^{2}-0.64 v_{1}-0.645=0\) Using the quadratic formula, $$ v_{1}=\frac{0.64 \pm \sqrt{(0.64)^{2}+5.16}}{4}=0.16 \pm 0.59 \mathrm{~m} / \mathrm{s} $$ from which \(v_{1}=0.75 \mathrm{~m} / \mathrm{s}\) or \(-0.43 \mathrm{~m} / \mathrm{s}\). Substitution back into \(\mathrm{Eq} .\) (1) gives \(v_{2}=-0.43 \mathrm{~m} / \mathrm{s}\) or \(0.75 \mathrm{~m} / \mathrm{s}\). Two choices for answers are available: \(\left(v_{1}=0.75 \mathrm{~m} / \mathrm{s}, v_{2}=-0.43 \mathrm{~m} / \mathrm{s}\right) \quad\) and \(\quad\left(v_{1}=-0.43 \mathrm{~m} / \mathrm{s}, v_{2}=0.75 \mathrm{~m} / \mathrm{s} \mathrm{s}\right.\) We must discard the first choice because it implies that the balls continue on unchanged; that is to say, no collision occurred. The correct answer is therefore \(v_{1}=-0.43 \mathrm{~m} / \mathrm{s}\) and \(v_{2}=0.75 \mathrm{~m} / \mathrm{s}\), which tells us that in a perfectly elastic, head-on collision between equal masses, the two bodies simply exchange velocities. Hence, \(\overrightarrow{\mathbf{v}}_{1}=0.43 \mathrm{~m} / \mathrm{s}-\) WEST and \(\overrightarrow{\mathbf{v}}_{2}=0.75 \mathrm{~m} / \mathrm{s}\) - EAST. Alternative Method If we recall that \(e=1\) for a perfectly elastic head-on collision, then $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad 1=\frac{v_{2}-v_{1}}{(0.75 \mathrm{~m} / \mathrm{s})-(20.43 \mathrm{~m} / \mathrm{s})} $$ which gives $$ v_{2}-v_{1}=1.18 \mathrm{~m} / \mathrm{s} $$ Equations (1) and (3) determine \(v_{1}\) and \(v_{2}\) uniquely.

Step by step solution

01

Understand Conservation of Momentum

In a perfectly elastic collision, momentum is conserved. So, we set up the momentum conservation equation: \[m(0.75 \, \text{m/s}) + m(-0.43 \, \text{m/s}) = m v_1 + m v_2\],which simplifies to:\[0.32 \, \text{m/s} = v_1 + v_2\].

02

Apply Conservation of Kinetic Energy

Kinetic energy is also conserved in a perfectly elastic collision. The equation for conservation of kinetic energy is:\[\frac{1}{2} m (0.75)^2 + \frac{1}{2} m (0.43)^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2\],which simplifies to:\[0.747 = v_1^2 + v_2^2\].

03

Solve for One Variable

From the momentum equation, solve for one of the velocities in terms of the other. Let's solve for \(v_2\):\[v_2 = 0.32 - v_1\].

04

Substitute and Solve the Quadratic Equation

Substitute \(v_2 = 0.32 - v_1\) into the kinetic energy equation:\[0.747 = (0.32 - v_1)^2 + v_1^2\].This simplifies to a quadratic equation:\[2v_1^2 - 0.64v_1 - 0.645 = 0\].

05

Use the Quadratic Formula

Apply the quadratic formula to solve for \(v_1\):\[v_1 = \frac{0.64 \pm \sqrt{(0.64)^2 + 4 \times 2 \times 0.645}}{4}\],Simplifying gives: \(v_1 = 0.75\, \text{m/s}\) or \(v_1 = -0.43\, \text{m/s}\).

06

Verify Velocity Pair

Substituting these values back into the equation for \(v_2\):- If \(v_1 = 0.75\, \text{m/s} \Rightarrow v_2 = -0.43\, \text{m/s}\), this means no collision, thus incorrect.- If \(v_1 = -0.43\, \text{m/s} \Rightarrow v_2 = 0.75\, \text{m/s}\), indicates a proper exchange of velocities.

07

Conclude with Correct Velocities

In an elastic collision involving identical masses, the bodies exchange velocities:- Ball 1 final velocity: \(v_1 = -0.43\, \text{m/s}\) (West).- Ball 2 final velocity: \(v_2 = 0.75\, \text{m/s}\) (East).

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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, described as the product of the mass and velocity of an object. In simple terms, it is * how much "motion" an object has * and is represented by the symbol \( p = m imes v \).During an elastic collision, especially a head-on one like in our exercise, the total momentum of the system (i.e., both objects involved in the collision) does not change. This is known as the conservation of momentum. It implies that the momentum before the collision equals the momentum after the collision.For our two identical balls:- Each ball has a mass \( m \), which remains constant.- The first ball moves eastward with velocity \( 0.75 \, \text{m/s} \).- The second one moves westward at \( 0.43 \, \text{m/s} \).Using conservation of momentum, the combined momentum can be observed as:\[ m(0.75 \, \text{m/s}) + m(-0.43 \, \text{m/s}) = m v_1 + m v_2 \]This results in:\[ 0.32 \, \text{m/s} = v_1 + v_2 \]This equation tells us that the sum of the velocities after the collision should equal the difference in initial velocities, reflecting how momentum is partitioned between the two balls.

Conservation of Kinetic Energy

Kinetic energy (KE) is the energy an object possesses because of its motion. It is derived using the equation:\[ \text{KE} = \frac{1}{2} m v^2 \]In perfectly elastic collisions, like the one in the exercise, both momentum and kinetic energy are conserved. This means the total kinetic energy before the collision is equal to the total kinetic energy after the collision.For our scenario:- Initial kinetic energy is calculated for both balls: \[ \text{KE}_{\text{before}} = \frac{1}{2} m (0.75)^2 + \frac{1}{2} m (0.43)^2 \]- Leading to the simplified equation: \[ 0.747 = v_1^2 + v_2^2 \]This equation lets us understand that although individual kinetic energies of the balls might change, the sum remains constant. This is a significant trait of elastic collisions, ensuring energy isn't lost but redistributed between the colliding entities.

Quadratic Equation

In solving the final velocities of the balls, we end up with a quadratic equation due to substituting one velocity into the kinetic energy equation. This is a typical approach when dealing with quadratic relationships:The basic form of a quadratic equation is:\[ ax^2 + bx + c = 0 \]For our problem:- After substituting \( v_2 = 0.32 - v_1 \),- The kinetic energy equation simplifies to: \[ 2v_1^2 - 0.64v_1 - 0.645 = 0 \]Using the quadratic formula given by:\[ v_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]we solve for \( v_1 \). Applying values:- \( a = 2 \),- \( b = -0.64 \),- \( c = -0.645 \).Calculation yields:- \( v_1 = 0.75 \, \text{m/s} \) or \( -0.43 \, \text{m/s} \)Using these potential solutions and confirming with momentum conservation helps verify the correct exchange of velocities, characteristic of elastic collisions.