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(II) A 0.450-kg hockey puck, moving east with a speed of 5.80 m/s, has a head- on collision with a 0.900-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each puck after the collision?

Short Answer

Expert verified
Puck 1 moves west at 1.93 m/s; puck 2 moves east at 3.87 m/s.

Step by step solution

01

Identify Conservation Laws Involved

In a perfectly elastic collision, both momentum and kinetic energy are conserved. Our approach will involve using these conservation laws to determine the speeds and directions of the pucks after the collision.
02

Apply Conservation of Momentum

The total momentum before and after the collision must be the same. The equation is \( m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \), where \( m_1 = 0.450\, \text{kg} \), \( v_{1i} = 5.80\, \text{m/s} \), \( m_2 = 0.900\, \text{kg} \) and \( v_{2i} = 0 \). Substitute the values to get \( 0.450 \times 5.80 + 0.900 \times 0 = 0.450v_{1f} + 0.900v_{2f} \).
03

Apply Conservation of Kinetic Energy

For a perfectly elastic collision, the kinetic energy before and after the collision is the same. The equation is \( \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \). Substitute the values: \( \frac{1}{2}\times 0.450 \times 5.80^2 + 0 = 0.450\frac{1}{2}v_{1f}^2 + 0.900\frac{1}{2}v_{2f}^2 \).
04

Solve the Momentum Equation for One Variable

Solve the conservation of momentum equation from Step 2 for one of the final velocities: \( v_{1f} = \frac{5.80 \times 0.450 - 0.900v_{2f}}{0.450} \).
05

Substitute into Kinetic Energy Equation

Substitute \( v_{1f} \) from Step 4 into the conservation of kinetic energy equation from Step 3. This will result in a quadratic equation in terms of \( v_{2f} \). Solve this quadratic equation to find \( v_{2f} \).
06

Calculate Final Velocities

With \( v_{2f} \) known from Step 5, use the equation from Step 4 to calculate \( v_{1f} \). These calculated speeds will provide the final velocities of the two pucks.
07

Analyze Directions

Since the pucks collide head-on and due to the symmetry of the problem, the first puck will travel in the opposite direction after the collision, while the second puck will move in the same direction as the first puck's initial direction.

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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 cornerstone principle that states the total momentum of an isolated system remains constant if no external forces are acting on it. In the context of collisions, the law of conservation of momentum can be expressed as the total momentum before the collision equaling the total momentum after the collision.

Here's how it works in our hockey puck problem: initially, we have a system with two pucks, one moving and one at rest. The formula involved is:

\[ m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \]

This equation shows that the sum of the initial momenta (from the moving puck) must equal the sum of the final momenta (from both pucks after the collision). By plugging in the specific values from our problem, we derive the relationship between the final velocities of these pucks.
  • This principle helps establish one of the equations used to solve for the final velocities of the colliding pucks.
  • Remember that each object's momentum is calculated by multiplying its mass by its velocity.
  • Conservation of momentum applies perfectly since we assume no external forces are interfering with the pucks' motion.
Conservation of Kinetic Energy
In an elastic collision, not only is momentum conserved, but kinetic energy is conserved as well. Kinetic energy is the energy an object has due to its motion, and for elastic collisions, the total kinetic energy before and after the collision remains constant.

This is mathematically expressed by the equation:

\[ \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \]

Within this framework, each puck's kinetic energy before the collision is compared with the total kinetic energy after the collision. The equation above allows us to determine another relationship between the final velocities of the pucks.
  • Conservation of kinetic energy is critical in perfectly elastic collisions, where no energy is lost to deformation or heat.
  • Kinetic energy depends on both mass and the square of velocity, so the impact of speed is very significant.
  • This principle gives us the second essential equation needed to solve our problem.
Final Velocities
The final velocities of objects involved in an elastic collision are determined by solving the combined systems of equations derived from conserving both momentum and kinetic energy.

**Interpreting the Outcome**
- For our hockey puck scenario, solving these equations reveals that the initially moving puck will reverse its direction after the collision while the initially stationary puck will move in the original direction of the first puck.

These outcomes can be intuitively understood when considering that the momentum and kinetic energy constraints result in a transfer of energy and momentum between the colliding pucks.
  • The direction of motion after the collision is predictable due to the symmetrical head-on interaction.
  • The mathematical calculations ensure that neither energy nor momentum is "lost" in any form.
Solving for the final velocities involved using algebra and can also involve quadratic equations, depending on the initial setup. Students may need to rearrange terms and substitute known values to unravel the problem correctly.

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Most popular questions from this chapter

(II) A 980-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.6 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact. What was that speed?

(III) An atomic nucleus of mass \(m\) traveling with speed \(\upsilon\) collides elastically with a target particle of mass \(2m\) (initially at rest) and is scattered at 90\(^{\circ}\). \((a)\) At what angle does the target particle move after the collision? \((b)\) What are the final speeds of the two particles? \((c)\) What fraction of the initial kinetic energy is transferred to the target particle?

(II) Billiard ball A of mass \(m_A = 0.120 kg\) moving with speed \(v_A = 2.80 m/s\) strikes ball B, initially at rest, of mass \(m_B = 0.140 kg\). As a result of the collision, ball A is deflected off at an angle of 30.0\(^{\circ}\) with a speed \(\upsilon'_{A} = 2.10 m/s\). \((a)\) Taking the \(\chi\) axis to be the original direction of motion of ball A, write down the equations expressing the conservation of momentum for the components in the \(\chi\) and \(y\) directions separately. (b) Solve these equations for the speed \(\upsilon'_{B}\), and angle \(\theta'_{B}\), of ball B after the collision. Do not assume the collision is elastic.

(II) The masses of the Earth and Moon are \(5.98 \times 10^{24} kg\) and \(7.35 \times 10^{22} kg\), respectively, and their centers are separated by \(3.84 \times 10^8 m\). \((a)\) Where is the \(_{CM}\) of the Earth-Moon system located? \((b)\) What can you say about the motion of the Earth-Moon system about the Sun, and of the Earth and Moon separately about the Sun?

Two asteroids strike head-on: before the collision, asteroid A \((m_A = 7.5 \times 10^{12} kg)\) has velocity 3.3 km/s and asteroid B \((m_B = 1.45 \times 10^{13} kg)\) has velocity 1.4 km/s in the opposite direction. If the asteroids stick together, what is the velocity (magnitude and direction) of the new asteroid after the collision?

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