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Chapter 7: Problem 60

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 \(B\) ?

Short Answer

Expert verified
The final velocity of B is \(\vec{v}_{B2} = 10 \ \text{m/s} \, \hat{\mathrm{i}} + 15 \ \text{m/s} \, \hat{\mathrm{j}}\)

Step by step solution

01

- List the given information

Mass of both bodies: \(m_A = 2.0 \ \text{kg}\) and\( m_B = 2.0 \ \text{kg}\) Initial velocities: \(\vec{v}_{A1} = 15 \ \text{m/s} \, \hat{\mathrm{i}} + 30 \ \text{m/s} \, \hat{\mathrm{j}}\) and \(\vec{v}_{B1} = -10 \ \text{m/s} \, \hat{\mathrm{i}} + 5 \ \text{m/s} \, \hat{\mathrm{j}}\)Final velocity of A: \(\vec{v}_{A2} = -5 \ \text{m/s} \, \hat{\mathrm{i}} + 20 \ \text{m/s} \, \hat{\mathrm{j}}\)We are to find the final velocity of B: \(\vec{v}_{B2}\)
02

- Apply conservation of momentum

Since no external forces act on the system, the total momentum before collision equals the total momentum after collision. \[ m_A \vec{v}_{A1} + m_B \vec{v}_{B1} = m_A \vec{v}_{A2} + m_B \vec{v}_{B2} \]
03

- Write the momentum equation in vector form

Plug in the known values into the conservation of momentum equation. \[ 2.0 \cdot (15 \, \hat{\mathrm{i}} + 30 \, \hat{\mathrm{j}}) + 2.0 \cdot (-10 \, \hat{\mathrm{i}} + 5 \, \hat{\mathrm{j}}) = 2.0 \cdot (-5 \, \hat{\mathrm{i}} + 20 \, \hat{\mathrm{j}}) + 2.0 \vec{v}_{B2} \]
04

- Simplify the equation

Combine and simplify terms to isolate \(\vec{v}_{B2}\).\[ (30 \ \text{m/s} \, \hat{\mathrm{i}} + 60 \ \text{m/s}\ \, \hat{\mathrm{j}}) + (-20 \ \text{m/s} \, \hat{\mathrm{i}} + 10 \ \text{m/s} \, \hat{\mathrm{j}}) = (-10 \ \text{m/s} \, \hat{\mathrm{i}} + 40 \ \text{m/s} \, \hat{\mathrm{j}}) + 2.0 \vec{v}_{B2}\] \[ 10 \ \text{m/s} \, \hat{\mathrm{i}} + 70 \ \text{m/s} \, \hat{\mathrm{j}} = -10 \ \text{m/s} \, \hat{\mathrm{i}} + 40 \ \text{m/s} \, \hat{\mathrm{j}} + 2.0 \vec{v}_{B2} \]
05

- Solve for \(\vec{v}_{B2}\)

Move terms to isolate \(\vec{v}_{B2}\). \[ 10 \ \text{m/s} \, \hat{\mathrm{i}} + 70 \ \text{m/s} \, \hat{\mathrm{j}} + 10 \ \text{m/s} \, \hat{\mathrm{i}} - 40 \ \text{m/s} \, \hat{\mathrm{j}} = 2 \vec{v}_{B2} \]\[ 20 \ \text{m/s} \, \hat{\mathrm{i}} + 30 \ \text{m/s} \, \hat{\mathrm{j}} = 2 \vec{v}_{B2} \]Divide by 2: \[ \vec{v}_{B2} = 10 \ \text{m/s} \, \hat{\mathrm{i}} + 15 \ \text{m/s} \, \hat{\mathrm{j}} \]

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Key Concepts

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

Collision Physics
In physics, a collision refers to an event where two or more bodies exert forces on each other for a relatively short time. Understanding collisions is crucial in studying momentum and energy conservation. Collisions can be categorized as elastic or inelastic.
Elastic collisions are those where the total kinetic energy of the system is conserved, meaning no energy is lost to heat, sound, or deformation. Inelastic collisions, on the other hand, involve some loss of kinetic energy.
To solve problems involving collisions, we often invoke the law of conservation of momentum, which states that if no external forces act on a system, the total momentum before the collision is equal to the total momentum after the collision.
Vector Quantities
Velocity and momentum are both vector quantities, which means they have both magnitude and direction. In our example, the velocities of bodies A and B are given in vector forms, including the unit vectors \( \hat{\mathrm{i}} \) and \( \hat{\mathrm{j}} \).
The unit vector \( \hat{\mathrm{i}} \) represents motion in the horizontal direction (x-axis), while the unit vector \( \hat{\mathrm{j}} \) represents motion in the vertical direction (y-axis). Calculating the final velocity of body B involves breaking down these vectors and understanding how they interact.
When handling vectors, we add or subtract the respective components separately. This means we deal with the \( \hat{\mathrm{i}} \)-components and the \( \hat{\mathrm{j}} \)-components individually before combining them to get the final vector.
Momentum
Momentum is defined as the product of an object's mass and its velocity, represented mathematically as \( \vec{p} = m \vec{v} \). Being a vector quantity, momentum considers both magnitude and direction.
The law of conservation of momentum tells us that the total momentum of a system remains constant if no external forces are acting on it. This principle is pivotal when analyzing collisions.
For our collision problem, we applied the conservation of momentum by setting the total initial momentum equal to the total final momentum. This allowed us to derive the final velocity of body B after the collision.
Here is the detailed equation we used to solve it: \[ m_A \vec{v}_{A1} + m_B \vec{v}_{B1} = m_A \vec{v}_{A2} + m_B \vec{v}_{B2} \]
Such equations can be simplified by plugging in known values, solving component-wise, and eventually isolating and solving for the unknown.

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Most popular questions from this chapter

A cue stick strikes a stationary pool ball, with an average force of \(50 \mathrm{~N}\) over a time of \(10 \mathrm{~ms}\). If the ball has mass \(0.20 \mathrm{~kg}\), what speed does it have just after impact?

A \(0.30 \mathrm{~kg}\) softball has a velocity of \(15 \mathrm{~m} / \mathrm{s}\) at an angle of \(35^{\circ}\) below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with a velocity of (a) \(20 \mathrm{~m} / \mathrm{s}\), vertically downward and (b) \(20 \mathrm{~m} / \mathrm{s}\), horizontally away from the batter and back toward the pitcher?

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with its original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball.

It is well known that bullets and other missiles fired at Superman simply bounce off his chest (Fig.7-22). Suppose that a gangster sprays Superman's chest with \(3 \mathrm{~g}\) bullets at the rate of 100 bullets/min, and the speed of each bullet is \(500 \mathrm{~m} / \mathrm{s}\). Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets?

When jumping straight down, you can be seriously injured if you land stiff- legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. Suppose you have a mass \(m\) and you jump off a wall of height \(h\). (a) Use what you learned about constant acceleration motion to find the speed with which you hit the ground. Assume you simply step off the wall, so your initial \(y\) velocity is zero. Ignore air resistance. (Express your answer in terms of the symbols given.) (b) Suppose that the time interval starting when your feet first touch the ground until you stop is \(\Delta t .\) Calculate the (average) net force acting on you during that interval. (Again, express your answer in terms of the symbols given.) (c) Suppose \(h=1 \mathrm{~m}\). If you land stiff-legged, the time it takes you to stop may be as short as \(2 \mathrm{~ms}\), whereas if you bend your knees, it might be as long as \(0.1\) s. Calculate the average net force that would act on you in the two cases. (d) The net force on you while you are stopping includes both the force of gravity and the force of the ground pushing up. Which of these forces do you think does you the injury? Explain your reasoning. (e) For the two cases in part (c), calculate the upward force the ground exerts on you.
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