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Chapter 6: Problem 43

Two ice skaters not paying attention collide in a completely inelastic collision. Prior to the collision, skater 1 , with a mass of \(60 \mathrm{~kg},\) has a velocity of \(5.0 \mathrm{~km} / \mathrm{h}\) eastward, and moves at a right angle to skater \(2,\) who has a mass of \(75 \mathrm{~kg}\) and a velocity of \(7.5 \mathrm{~km} / \mathrm{h}\) southward. What is the velocity of the skaters after collision?

Short Answer

Expert verified
The final velocity is 1.31 m/s at an angle of 61.7掳 south of east.

Step by step solution

01

Convert Units

First, convert the velocities from kilometers per hour (km/h) to meters per second (m/s). The conversion factor is \(1 \text{ km/h} =\frac{1}{3.6} \text{ m/s}\). \[\text{Velocity of Skater 1} = 5.0 \text{ km/h} \times \frac{1}{3.6} = 1.39 \text{ m/s}\] \[\text{Velocity of Skater 2} = 7.5 \text{ km/h} \times \frac{1}{3.6} = 2.08 \text{ m/s}\]
02

Apply Conservation of Momentum

Since the collision is completely inelastic, we apply the conservation of momentum principle in both the x (east-west) and y (north-south) directions. The total momentum before and after the collision will be equal. \[m_1v_{1x} + m_2v_{2x} = (m_1 + m_2)v_{fx}\] \[m_1v_{1y} + m_2v_{2y} = (m_1 + m_2)v_{fy}\] In this problem, slightly note that skater 1's motion affects the x-direction and skater 2 affects the y-direction.
03

Calculate Momentum in x-direction

Since only skater 1 has initial velocity in the x-direction: \(v_{2x} = 0\). \[m_1v_{1x} = (m_1 + m_2)v_{fx}\] \[(60 \text{ kg})(1.39 \text{ m/s}) = (60 + 75)v_{fx}\] \[83.4 = 135v_{fx}\] \[v_{fx} = \frac{83.4}{135} = 0.618 \text{ m/s}\]
04

Calculate Momentum in y-direction

Since only skater 2 has initial velocity in the y-direction: \(v_{1y} = 0\). \[m_2v_{2y} = (m_1 + m_2)v_{fy}\] \[(75 \text{ kg})(2.08 \text{ m/s}) = (60 + 75)v_{fy}\] \[156 = 135v_{fy}\] \[v_{fy} = \frac{156}{135} = 1.156 \text{ m/s}\]
05

Determine Final Velocity

Combining velocities in the x and y directions using the Pythagorean theorem, the magnitude of the final velocity \(v_f\) is: \[v_f = \sqrt{v_{fx}^2 + v_{fy}^2}\] \[v_f = \sqrt{(0.618)^2 + (1.156)^2}\] \[v_f = \sqrt{0.382 + 1.336}\] \[v_f = \sqrt{1.718} = 1.31 \text{ m/s}\]
06

Determine the Direction of Velocity

The direction \(\theta\) (south of east) of the final velocity can be determined using the tangent function: \[\theta = \tan^{-1}\left(\frac{v_{fy}}{v_{fx}}\right)\] \[\theta = \tan^{-1}\left(\frac{1.156}{0.618}\right)\] \[\theta = \tan^{-1}(1.87)\] \[\theta \approx 61.7^{\circ}\]

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Key Concepts

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

Conservation of Momentum
In an inelastic collision, such as the one where two ice skaters collide, the **conservation of momentum** is a crucial concept. Momentum, which is the product of mass and velocity, remains constant in isolated systems unless acted upon by external forces. This means that the total momentum before the collision is equal to the total momentum after the collision. The equation representing this principle is:
  • For the x-direction: \( m_1v_{1x} + m_2v_{2x} = (m_1 + m_2)v_{fx} \)
  • For the y-direction: \( m_1v_{1y} + m_2v_{2y} = (m_1 + m_2)v_{fy} \)
Applying this principle allows us to predict the final velocities in both the x and y directions after the collision. It鈥檚 important to pay attention to the direction of the velocities, as momenta along different directions should be handled separately.
Momentum in x and y Directions
Calculating **momentum in the x and y directions** helps in assessing how each component of velocity changes as a result of a collision. Momentum is a vector quantity, meaning it has both magnitude and direction. Here鈥檚 how it works in different directions:
  • **x-direction (east-west):** Skater 1 initially moves east, contributing to the x-direction momentum. Skater 2 has no velocity in the x-direction, so their contribution to momentum here is zero. Thus, momentum before the collision is calculated using skater 1's momentum.

  • **y-direction (north-south):** Skater 2 initially moves south, contributing to the y-direction momentum. Skater 1 has no velocity in the y-direction, so their contribution here is zero. Hence, y-direction momentum calculation uses data from skater 2.
This division into x and y components makes solving two-dimensional problems simpler.
Velocity Conversion
When dealing with velocity, unit conversion is often necessary, especially when working between different measurement systems. Here, **velocity conversion** refers to changing units from kilometers per hour (km/h) to meters per second (m/s) for compatibility with standard equations in physics. The conversion factor for this is:
  • 1 km/h = \( \frac{1}{3.6} \) m/s
Using this, the velocities of the skaters are converted to:
  • Skater 1: 5.0 km/h is converted to 1.39 m/s
  • Skater 2: 7.5 km/h is converted to 2.08 m/s
Converting the velocities ensures that calculations are consistent with the International System of Units (SI), streamlining mathematical operations.
Direction of Velocity
Understanding the **direction of velocity** is essential in determining the precise movement of objects after they collide. Post-collision, the velocity direction is often expressed concerning a reference point, in this instance south of east. This compass-based expression provides a clear geographical interpretation.
For our problem, the direction is calculated by employing the tangent function with the ratios of the speeds in the y and x directions, \( \theta = \tan^{-1}\left(\frac{v_{fy}}{v_{fx}}\right) \):
  • \( \theta = \tan^{-1}\left(\frac{1.156}{0.618}\right) \)

  • This results in an approximate angle of 61.7 degrees south of east
This method involves trigonometry and emphasizes that direction is relative, altering based on the initial motion vectors.

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Most popular questions from this chapter

A cherry bomb explodes into three pieces of equal mass. One piece has an initial velocity of \(10 \mathrm{~m} / \mathrm{s} \hat{\mathrm{x}}\). Another piece has an initial velocity of \(6.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{x}}-3.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{y}} .\) What is the velocity of the third piece?

A 15.0 - \(\mathrm{g}\) rubber bullet hits a wall with a speed of \(150 \mathrm{~m} / \mathrm{s}\). If the bullet bounces straight back with a speed of \(120 \mathrm{~m} / \mathrm{s}\), what is the change in momentum of the bullet?

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Two balls of equal mass \((0.50 \mathrm{~kg})\) approach the origin along the positive \(x\) - and \(y\) -axes at the same speed \((3.3 \mathrm{~m} / \mathrm{s})\). (a) What is the total momentum of the system? (b) Will the balls necessarily collide at the origin? What is the total momentum of the system after both balls have passed through the origin?

(a) The center of mass of a system consisting of two 0.10-kg particles is located at the origin. If one of the particles is at \((0,0.45 \mathrm{~m}),\) where is the other? \((\mathrm{b})\) If the masses are moved so their center of mass is located at \((0.25 \mathrm{~m}\), \(0.15 \mathrm{~m}\) ), can you tell where the particles are located?
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