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In waterskiing, a "garage sale" occurs when a skier loses control and falls and waterskis fly in different directions. In one particular incident, a novice skier was skimming across the surface of the water at \(22.0 \mathrm{~m} / \mathrm{s}\) when he lost control. One ski, with a mass of \(1.50 \mathrm{~kg},\) flew off at an angle of \(12.0^{\circ}\) to the left of the initial direction of the skier with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). The other identical ski flew from the crash at an angle of \(5.00^{\circ}\) to the right with a speed of \(21.0 \mathrm{~m} / \mathrm{s} .\) What was the velocity of the \(61.0-\mathrm{kg}\) skier? Give a speed and a direction relative to the initial velocity vector.

Short Answer

Expert verified
Answer: To find the final velocity of the skier, first calculate the initial momentum of the skier and the momenta of the flying water skis using the given information. Then, apply the conservation of momentum principle to find the final momentum of the skier. Divide the final momentum of the skier by the skier's mass to get the final velocity. The final speed and direction can then be calculated using the magnitudes and inverse tangent function, respectively.

Step by step solution

01

Identify the initial momentum of the skier

Before falling, the skier is moving at a velocity of \(22.0 m/s\). The mass of the skier is \(61.0 kg\). Therefore, we can calculate the initial momentum of the skier as follows: Initial momentum of skier = mass * velocity = \(61.0 kg \times 22.0 m/s = 1342 kg\cdot m/s\)
02

Compute the momentum of water skis

Let's compute the momenta of the flying water skis: Both skis have a mass of \(1.50 kg\). The first ski is flying at \(25.0 m/s\) at an angle of \(12.0^\circ\) to the left and the second ski is flying at \(21.0 m/s\) at an angle of \(5.0^\circ\) to the right of the initial direction of the skier. To find the momentum of each ski, we have to resolve the velocity vector of each ski into its horizontal and vertical components. - For the first ski: Horizontal component = \(25 \cos(12^\circ)\) Vertical component = \(25 \sin(12^\circ)\) - For the second ski: Horizontal component = \(21 \cos(5^\circ)\) Vertical component = \(21 \sin(5^\circ)\) The momentum of each ski can then be computed as follows: - Momentum of the first ski = \((1.5 kg)(25\cos(12^\circ)\hat{i} + 25\sin(12^\circ)\hat{j})\) - Momentum of the second ski = \((1.5 kg)(21\cos(5^\circ)\hat{i} + 21\sin(5^\circ)\hat{j})\)
03

Apply the conservation of momentum

The total initial momentum equals the total final momentum. That is, the initial momentum of the skier is equal to the sum of the final momentum of the skier and the momenta of the flying water skis. Therefore, the final momentum of the skier can be calculated as follows: Final momentum of skier = Initial momentum of skier - (momentum of first ski + momentum of second ski) Final momentum of skier = \(1342\hat{i} kg\cdot m/s - [(1.5 kg)(25\cos(12^\circ)\hat{i} + 25\sin(12^\circ)\hat{j}) + (1.5 kg)(21\cos(5^\circ)\hat{i} + 21\sin(5^\circ)\hat{j})]\) Now, calculate the horizontal and vertical components of the final momentum of the skier.
04

Calculate the final velocity of the skier

Now that we have the final momentum of the skier, we can find the final velocity of the skier by dividing the final momentum by the mass of the skier: Final velocity of skier = \(\frac{Final~momentum~of~skier}{mass~of~skier}\) = \(\frac{Final~momentum~of~skier}{61.0 kg}\) To find the final speed of the skier, we will find the magnitude of the final velocity vector: Speed of the skier = \(|\textrm{Final velocity of skier}| = \sqrt{(Final~Horizontal~Component)^2 + (Final~Vertical~Component)^2}\) To find the direction of the skier with respect to the initial velocity vector, we use the inverse tangent (arctan) function: Direction of the skier = \(tan^{-1}\left(\frac{Final~Vertical~Component}{Final~Horizontal~Component}\right)\) Calculate the final speed and direction of the skier using the computed values.

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Key Concepts

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

Vector Components
Understanding vector components is crucial in physics, particularly when analyzing motion in different directions. A vector has both magnitude and direction, making it essential to break it down into more manageable parts for calculations.

In our example, we calculated the motion of the skis after the skier fell. Each ski flies off at an angle to the initial direction of travel. To analyze these motions effectively, we resolve the velocity of each ski into horizontal (x-component) and vertical (y-component) components. This is done using trigonometric functions, sine and cosine:

  • Horizontal component is given by multiplying the speed by the cosine of the angle ( \( \text{velocity} \times \cos(\theta) \))
  • Vertical component is found using the sine of the angle ( \( \text{velocity} \times \sin(\theta) \))
By resolving vectors into components, it is easier to work with them in equations and match them up in both magnitude and direction.
Problem Solving
Problem-solving in physics often involves creative thinking and careful calculations. The key to success usually lies in clearly understanding the problem and applying the relevant principles effectively.

In the given exercise, solving the problem involves several steps:

  • Calculating the initial momentum of the skier, which is straightforward (mass multiplied by velocity).
  • Finding momentum vectors of the skis by resolving their velocities into components and multiplying by mass.
  • Using the conservation of momentum, which states that the total momentum before an event is equal to the total momentum after it, to find the missing information regarding the skier's velocity after the skis fly off.
A robust approach can significantly ease complex problems: break down the information, consider each piece separately, and think logically through each step.
Physics Applications
Physics has real-world applications which, when seen through exercises like this, become more tangible and understandable. The conservation of momentum is one of the fundamental principles applicable to various phenomena, from vehicle collisions to sports incidents like water skiing crashes.

This principle allows us to predict the outcomes after events like our skier's fall. Recognizing that total momentum must remain constant tells us how different parts of the system—here, the skier and skis—move after a crash. Additionally, understanding vector components is necessary for accurately describing physical phenomena which occur in multiple directions.

By studying physics through practical applications, we gain valuable insights into how concepts like momentum and vector components interact in scenarios encountered in real life, thus improving both our problem-solving skills and our grasp of the physical world.

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Most popular questions from this chapter

A small car of mass 1000 . kg traveling at a speed of \(33.0 \mathrm{~m} / \mathrm{s}\) collides head on with a large car of mass \(3000 \mathrm{~kg}\) traveling in the opposite direction at a speed of \(30.0 \mathrm{~m} / \mathrm{s}\). The two cars stick together. The duration of the collision is \(100 . \mathrm{ms}\). What acceleration (in \(g\) ) do the occupants of the small car experience? What acceleration (in \(g\) ) do the occupants of the large car experience?

A bullet with mass \(35.5 \mathrm{~g}\) is shot horizontally from a gun. The bullet embeds in a 5.90 -kg block of wood that is suspended by strings. The combined mass swings upward, gaining a height of \(12.85 \mathrm{~cm}\). What was the speed of the bullet as it left the gun? (Air resistance can be ignored here.)

Bats are extremely adept at catching insects in midair. If a 50.0-g bat flying in one direction at \(8.00 \mathrm{~m} / \mathrm{s}\) catches a \(5.00-\mathrm{g}\) insect flying in the opposite direction at \(6.00 \mathrm{~m} / \mathrm{s}\), what is the speed of the bat immediately after catching the insect?

A billiard ball of mass \(m=0.250 \mathrm{~kg}\) hits the cushion of a billiard table at an angle of \(\theta_{1}=60.0^{\circ}\) at a speed of \(v_{1}=27.0 \mathrm{~m} / \mathrm{s}\) It bounces off at an angle of \(\theta_{2}=71.0^{\circ}\) and a speed of \(v_{2}=10.0 \mathrm{~m} / \mathrm{s}\). a) What is the magnitude of the change in momentum of the billiard ball? b) In which direction does the change of momentum vector point?

An uncovered hopper car from a freight train rolls without friction or air resistance along a level track at a constant speed of \(6.70 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. The mass of the car is \(1.18 \cdot 10^{5} \mathrm{~kg}\). a) As the car rolls, a monsoon rainstorm begins, and the car begins to collect water in its hopper (see the figure). What is the speed of the car after \(1.62 \cdot 10^{4} \mathrm{~kg}\) of water collects in the car's hopper? Assume that the rain is falling vertically in the negative \(y\) -direction. b) The rain stops, and a valve at the bottom of the hopper is opened to release the water. The speed of the car when the valve is opened is again \(6.70 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction (see the figure). The water drains out vertically in the negative \(y\) -direction. What is the speed of the car after all the water has drained out?

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