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(II) A \(0.450-\mathrm{kg}\) hockey puck, moving east with a speed of \(4.80 \mathrm{~m} / \mathrm{s},\) has a head-on collision with a \(0.900-\mathrm{kg}\) puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each object after the collision?

Short Answer

Expert verified
Puck 1 moves at 1.60 m/s west, and puck 2 moves at 3.20 m/s east after the collision.

Step by step solution

01

Identify Known Values

First, let's identify all the known values from the problem. We have two pucks:For puck 1:- Mass, \( m_1 = 0.450 \ \mathrm{kg} \)- Initial velocity, \( u_1 = 4.80 \ \mathrm{m/s} \) (moving east)For puck 2:- Mass, \( m_2 = 0.900 \ \mathrm{kg} \)- Initial velocity, \( u_2 = 0 \ \mathrm{m/s} \) (at rest)We're tasked with finding the final velocities \( v_1 \) and \( v_2 \) after the collision.
02

Apply Conservation of Momentum

In a perfectly elastic collision, both momentum and kinetic energy are conserved. First, apply the conservation of linear momentum:\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]Substitute the known values:\[ 0.450 \times 4.80 + 0.900 \times 0 = 0.450 \times v_1 + 0.900 \times v_2 \]Simplify:\[ 2.16 = 0.450v_1 + 0.900v_2 \](Equation 1)
03

Apply Conservation of Kinetic Energy

Since the collision is perfectly elastic, use the conservation of kinetic energy:\[ \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \]Substitute the known values:\[ \frac{1}{2} \times 0.450 \times (4.80)^2 + \frac{1}{2} \times 0.900 \times 0^2 = \frac{1}{2} \times 0.450 \times v_1^2 + \frac{1}{2} \times 0.900 \times v_2^2 \]Simplify and arrange:\[ 5.184 = 0.225v_1^2 + 0.450v_2^2 \](Equation 2)
04

Solve the System of Equations

We now have two equations:1. \[ 0.450v_1 + 0.900v_2 = 2.16 \]2. \[ 0.225v_1^2 + 0.450v_2^2 = 5.184 \] Let's solve these equations starting with Equation 1. Express \( v_1 \) in terms of \( v_2 \):\[ v_1 = \frac{2.16 - 0.900v_2}{0.450} \]Substitute this expression into Equation 2:\[ 0.225\left(\frac{2.16 - 0.900v_2}{0.450}\right)^2 + 0.450v_2^2 = 5.184 \]Solve this equation to find \( v_2 \). Once calculated, use the value of \( v_2 \) to find \( v_1 \) from the earlier expression. The solutions for \( v_1 \) and \( v_2 \) should satisfy both equations.
05

Calculate Final Velocities

Solve the simplified equation for \( v_2 \). After calculation:\( v_2 = 3.20 \ \mathrm{m/s} \) (east)Substitute \( v_2 \) back into the equation for \( v_1 \):\( v_1 = \frac{2.16 - 0.900 \times 3.20}{0.450} \)Calculate \( v_1 \):\( v_1 = 1.60 \ \mathrm{m/s} \) (west)

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Key Concepts

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

Conservation of Momentum
The conservation of momentum is a fundamental principle in physics that states that in a closed system with no external forces, the total momentum remains constant before and after a collision. This principle is extremely useful in analyzing collisions, such as the one between the two hockey pucks in our exercise.

In our problem, we have two pucks colliding head-on. The first puck, with a mass of 0.450 kg, moves at a velocity of 4.80 m/s. The second puck is initially at rest, with a mass of 0.900 kg. The law of conservation of momentum is represented mathematically as:
  • \[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
This equation allows us to relate the initial momenta of the pucks to their final momenta. By inputting the known values from our exercise, we solve for the unknown final velocities, \( v_1 \) and \( v_2 \). Using this principle, we preserve the total momentum of the system, ensuring we can accurately predict the pucks' final states post-collision.
Conservation of Kinetic Energy
In a perfectly elastic collision, not only is momentum conserved, but kinetic energy is also preserved. This means that the total kinetic energy of both objects before the collision is equal to the total kinetic energy after the collision.

The formula for conservation of kinetic energy in our problem is:
  • \[ \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \]
Applying this to our exercise, we calculate the kinetic energy of each puck. Initially, only the first puck possesses kinetic energy because the second puck is at rest. Solving this equation alongside our momentum conservation equation helps us find the final velocities, \( v_1 \) and \( v_2 \).

The calculated final velocities must satisfy both the momentum and energy equations simultaneously, as these are the hallmark conditions of an elastic collision.
Physics Problem-Solving
Tackling physics problems like this one involves a structured approach, often starting with identifying known and unknown quantities. In our case, we began by noting the masses and initial velocities of the two hockey pucks.

A systematic method includes:
  • Identifying the physical principles at play, such as the conservation laws involved in the collision.
  • Setting up the appropriate equations to represent these physical principles, as shown with our momentum and kinetic energy equations.
  • Solving the system of equations by substitution or elimination to find unknown variables, such as the final velocities here.
This process is critical in physics problem-solving. It allows students to translate real-world scenarios into mathematical formulations that can be solved to derive meaningful results.

By practicing this method, students build the analytical skills necessary to tackle more complex problems, providing a strong foundation for understanding physics.

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Most popular questions from this chapter

(III) A 3.0-kg block slides along a frictionless tabletop at \(8.0 \mathrm{~m} / \mathrm{s}\) toward a second block (at rest) of mass \(4.5 \mathrm{~kg} . \mathrm{A}\) coil spring, which obeys Hooke's law and has spring constant \(k=850 \mathrm{~N} / \mathrm{m},\) is attached to the second block in such a way that it will be compressed when struck by the moving block, Fig. \(9-40 .\) ( \(a\) ) What will be the maximum compression of the spring? (b) What will be the final velocities of the blocks after the collision? \((c)\) Is the collision elastic? Ignore the mass of the spring.

(II) A 145-g baseball, moving along the \(x\) axis with speed \(30.0 \mathrm{~m} / \mathrm{s}\), strikes a fence at a \(45^{\circ}\) angle and rebounds along the \(y\) axis with unchanged speed. Give its change in momentum using unit vector notation.

A \(4800-\mathrm{kg}\) open railroad car coasts along with a constant speed of \(8.60 \mathrm{~m} / \mathrm{s}\) on a level track. Snow begins to fall vertically and fills the car at a rate of \(3.80 \mathrm{~kg} / \mathrm{min}\). Ignoring friction with the tracks, what is the speed of the car after \(60.0 \mathrm{~min} ?\) (See Section 9-2.)

(III) Determine the \(\mathrm{CM}\) of a machine part that is a uniform cone of height \(h\) and radius \(R\), Fig. 9-46. [Hint: Divide the cone into an infinite number of disks of thickness \(d z,\) one of which is shown.

(III) A \(280-\mathrm{kg}\) flatcar 25 \(\mathrm{m}\) long is moving with a speed of 6.0 \(\mathrm{m} / \mathrm{s}\) along horizontal frictionless rails. A \(95-\mathrm{kg}\) worker starts walking from one end of the car to the other in the direction of motion, with speed 2.0 \(\mathrm{m} / \mathrm{s}\) with respect to the car. In the time it takes for him to reach the other end, how far has the flatcar moved?

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