/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 The windscreens of two motorcars... [FREE SOLUTION] | 91Ó°ÊÓ

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The windscreens of two motorcars are having slopes \(30^{\circ}\) and \(15^{\circ}\) respectively. At what ratio \(v_{1} / v_{2}\) of the velocities of cars will their drivers see the hailstones bounced by windscreen of their cars in the vertical direction? Assume hailstones are falling vertically.

Short Answer

Expert verified
The ratio is approximately 0.732.

Step by step solution

01

Understanding the Problem

We want to find the velocity ratio \( \frac{v_1}{v_2} \) such that the components of velocity of the hailstones along the slope of the windscreens make the hailstones appear to bounce vertically. For a hailstone to bounce vertically along a slope, the horizontal component of the car's velocity must cancel out the horizontal component of the hailstone's velocity relative to the earth along the slope.
02

Establishing the Components

Given the slopes of the windscreens are \(30^{\circ}\) and \(15^{\circ}\), the velocity of hailstones is purely vertical. The relative velocity along the slope for a car is \(v \sin \theta\) (horizontal component of hailstone) and \(v \cos \theta\) (vertical component of hailstone). The windscreen's angle leads the bounce to appear vertical in relation.
03

Condition for Bouncing Vertically

For the hailstones to appear to bounce vertically, the horizontal component of the car's velocity \(v\) must be equal to the horizontal component of the relative velocity of the hailstones. Thus, we have:\[ v_1 \sin 30^{\circ} = v_2 \sin 15^{\circ} \]
04

Solving the Equation

Substituting the values of the sine functions gives:\[ v_1 \times \frac{1}{2} = v_2 \times \frac{\sqrt{3} - 1}{2\sqrt{2}} \]Solving for the ratio \( \frac{v_1}{v_2} \), we find:\[ \frac{v_1}{v_2} = \frac{\sqrt{3} - 1}{\sqrt{2}} \times 2 \]
05

Simplifying the Ratio

We further simplify the expression:\[ \frac{v_1}{v_2} = \sqrt{3} - 1 \approx 0.732 \]

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Key Concepts

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

Vector Components
In the realm of physics, understanding vector components is vital, especially when dealing with relative motion. Vectors allow us to break down movements into smaller, understandable parts. For this problem, hailstones fall vertically, characterized by a downward vector. When these hailstones hit a sloped windscreen, their velocity can be dissected into two components: horizontal and vertical.
1. **Horizontal Component:** This is the part of the vector acting parallel to the slope of the windscreen. It can be calculated using the sine function, as in the expression: \(v \sin \theta\).
2. **Vertical Component:** Operating perpendicular to the slope, this is determined using the cosine function, represented by \(v \cos \theta\).

By analyzing these components, we can determine how the car's movement interacts with the falling hailstones. Specifically, modifying the car's velocity can align the horizontal component of its motion with the opposite horizontal component created by the slope. This creates the appearance of a vertical bounce.
Velocity Ratio
The velocity ratio \(\frac{v_1}{v_2}\) is a crucial part of understanding relative motion between two objects, here, two cars. This concept involves comparing the speeds of the cars, without delving into their absolute velocities, and makes it easier to understand the problem at hand.

When we express this ratio using the angles of the windscreens and the sine functions, we establish an equation.
  • For car 1: Velocity is expressed in terms of the \(30^\circ\) windscreen angle.
  • For car 2: Velocity is expressed in terms of the \(15^\circ\) windscreen angle.
This relative comparison through angles allows us to solve for the required speed relationship for the vertical bouncing appearance. Hence, \[ v_1 \sin 30^\circ = v_2 \sin 15^\circ \] is established to equate the horizontal components, leading to a calculated ratio that provides insight into the necessary speeds for both cars.
Trigonometric Functions
Trigonometric functions underpin the calculations in this problem scenario, measuring angles to resolve vector components. These functions—particularly sine and cosine—are used to dissect the movement of the hailstones and car's motion in relation to slope angles.

For a windscreen with a given slope angle \(\theta\):
  • The **sine function** (\(\sin\)) gives the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. Here, it helps break down the horizontal component of velocity.
  • The **cosine function** (\(\cos\)) provides the ratio of the adjacent side to the hypotenuse, assisting in understanding the vertical component.
These functions allow for precise analysis of the problem: equating the horizontal components of the velocities and thereby enabling the hailstones to appear to bounce vertically. By using specific trigonometric identities and solving these equations, one can simplify complex vector relationships into manageable terms, leading to the solution of the problem.

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Most popular questions from this chapter

Two balloons are simultaneously released from two buildings \(A\) and \(B\). Balloon from A rises with constant velocity \(10 \mathrm{~ms}^{-1}\), While the other one rises with constant velocity of \(20 \mathrm{~ms}^{-1}\). Due to wind the balloons gather horizontal velocity \(V_{x}=0.5 y\), where \(y\) is the height from the point of release. The buildings are at a distance of \(250 \mathrm{~m}\) and after some time \(t\) the balloons collide. Choose the correct option(s). (1) \(t=5 \mathrm{sec}\) (2) difference in height of buildings is \(100 \mathrm{~m}\) (3) difference in height of buildings is \(50 \mathrm{~m}\) (4) \(t=10 \mathrm{sec}\)

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A car is moving towards east with a speed of \(25 \mathrm{~km} \mathrm{~h}^{-1}\). To the driver of the car, a bus appears to move towards north with a speed of \(25 \sqrt{3} \mathrm{~km} \mathrm{~h}^{-1}\). What is the actual velocity of the bus? (1) \(50 \mathrm{~km} \mathrm{~h}^{-1}, 30^{\circ} \mathrm{E}\) of \(\mathrm{N}\) (2) \(50 \mathrm{~km} \mathrm{~h}^{-1}, 30^{\circ} \mathrm{N}\) of \(\mathrm{E}\) (3) \(25 \mathrm{~km} \mathrm{~h}^{-1}, 30^{\circ} \mathrm{E}\) of \(\mathrm{N}\) (4) \(25 \mathrm{~km} \mathrm{~h}^{-1}, 30^{\circ} \mathrm{N}\) of \(\mathrm{E}\)

Shots are fired simultaneously from the top and bottom of a vertical cliff with the elevation \(\alpha=30^{\circ}\), \(\beta=60^{\circ}\), respectively (figure). The shots strike an object simultaneously at the same point. If \(a=30 \sqrt{3} \mathrm{~m}\) is the horizontal distance of the object from the cliff, then the height \(h\) of the cliff is (1) \(30 \mathrm{~m}\) (2) \(45 \mathrm{~m}\) (3) \(60 \mathrm{~m}\) (4) \(90 \mathrm{~m}\)

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