/*! 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 9 Two inclined planes \(O A\) and ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 9

Two inclined planes \(O A\) and \(O B\) having inclination (with borizontal) \(30^{\circ}\) and \(60^{\circ}\), respectively, intersect each other at \(O\) as shown in figure. A particle is projected from point \(P\) with velocity \(u=10 \sqrt{3} \mathrm{~ms}^{-1}\) along a direction perpendicular to plane \(O A\). If the particle strikes plane \(O B\) perpendicularly at \(Q\), calculate The velocity with which particle strikes the plane \(O B\), (1) \(15 \mathrm{~ms}^{-1}\) (2) \(30 \mathrm{~ms}^{\prime}\) (3) \(20 \mathrm{~ms}^{-1}\) (4) \(10 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The velocity with which the particle strikes the plane \( OB \) is 10 \( \text{m/s} \).

Step by step solution

01

Analyze Initial Motion

The particle is projected from point \( P \) along the plane \( OA \) with a velocity \( u = 10 \sqrt{3} \text{ m/s} \), and it is perpendicular to the plane \( OA \). Since the plane is inclined at \( 30^{\circ} \) to the horizontal, the velocity \( u \) makes an angle of \( 60^{\circ} \) with the horizontal (complement of the inclination).
02

Resolve Initial Velocity

Resolve initial velocity \( u \) into horizontal and perpendicular components concerning the horizontal axis:- Horizontal component: \( u_x = 10 \sqrt{3} \cos(60^{\circ}) = 10 \sqrt{3} \times \frac{1}{2} = 5 \sqrt{3} \text{ m/s} \)- Perpendicular component: \( u_y = 10 \sqrt{3} \sin(60^{\circ}) = 10 \sqrt{3} \times \frac{\sqrt{3}}{2} = 15 \text{ m/s} \)
03

Determine the Time of Flight (t)

Since the particle strikes the plane \( OB \) perpendicularly and it is inclined at \( 60^{\circ} \) with the horizontal, the vertical motion affects the particle until it reaches \( B \). Given that the particle is initially projected perpendicular to \( OA \), it maintains constant horizontal velocity. To find the time to reach \( B \), consider that motion along \( OB \) does not involve vertical motion.
04

Calculate the Horizontal Distance (S)

As the particle strikes plane \( OB \) perpendicularly, it covers a horizontal distance determined by the change in its trajectory due to gravity.The horizontal distance \( S \) can be derived using: \[ S = u_x \times t \]where \( u_x = 5 \sqrt{3} \).
05

Calculate Final Velocity at Q

When the particle hits plane \( OB \) perpendicularly, only the velocity component normal to \( OB \) is considered. Since it strikes at \( OB \) perpendicularly, the velocity of the particle at \( Q \) along the direction of projection is:\[ v = 10 \text{ m/s} \]The normal component of the initial velocity that transfers to the motion is sustained through gravitational effect adjustment following the projectile's path.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Inclined Planes
Inclined planes play a crucial role in physics problems, especially those involving projectile motion. When a surface is tilted, the gravitational force acting on an object must be split into components.
For this problem, there are two inclined planes: Plane OA inclined at 30掳 and Plane OB at 60掳.
This inclination affects the path and impact of projectiles. The forces on the object are divided into two directions:
  • Parallel to the inclined plane: affects acceleration but means no perpendicular motion occurs.
  • Perpendicular to the inclined plane: affects how the velocity changes when striking another plane.
Understanding how a projectile interacts with inclined planes helps in predicting and calculating its path as it travels across different inclinations.
Resolving Vectors
Resolving vectors is about breaking a velocity vector into perpendicular components. This helps in analyzing the projectile's journey step-by-step.
For the projectile in this problem, it starts with a velocity vector perpendicular to Plane OA.
We resolve this vector into components parallel and perpendicular to the horizontal plane:
  • Horizontal component: Calculated as \( u_x = u \cos(60^{\circ}) \), where \( u = 10 \sqrt{3} \text{ m/s} \). This component is \( 5 \sqrt{3} \text{ m/s} \).
  • Perpendicular component: Calculated as \( u_y = u \sin(60^{\circ}) \), resulting in \( 15 \text{ m/s} \).
These resolved components determine the motion and the way the projectile interacts with other planes, like Plane OB.
Perpendicular Projection
When a projectile strikes a surface perpendicularly, it simplifies calculations as it reduces the components that need to be considered.
For Plane OB, the projectile strikes it perpendicularly, meaning that only the perpendicular component of the initial velocity is relevant.
Important points about perpendicular projection:
  • Only the component of velocity normal to the impact surface affects the motion at the point of impact.
  • The calculated final velocity at the strike is simplified to just being the normal component, ensuring easy calculation.
In this specific problem, the projectile's perpendicular velocity at Plane OB led to a straightforward determination of the impact speed as \( 10 \text{ m/s} \), utilizing the sustained perpendicular motion due to gravity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The height \(y\) and the distance \(x\) along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by \(y=\left(8 t-5 t^{2}\right) \mathrm{m}\) and \(x=6 t \mathrm{~m}\), where \(t\) is in seconds. The velocity with which the projectile is projected at \(t=0\) is (1) \(8 \mathrm{~m} \mathrm{~s}^{-1}\) (2) \(6 \mathrm{~ms}^{-1}\) (3) \(10 \mathrm{~ms}^{-1}\) (4) Not obtainable from the data

A person sitting in the rear end of a compartment throws a ball towards the front end. The ball follows a parabolic path. The train is moving with the uniform velocity of \(20 \mathrm{~ms}^{-1}\). A person standing outside on the ground also observes the ball. How will the maximum heights \(\left(h_{m}\right)\) attained and the ranges \((R)\) seen by the thrower and the outside observer compare each other? (1) Same \(h_{m}\), different \(R\) (2) same \(\mathrm{h}_{\mathrm{m}}\), and \(\mathrm{R}\) (3) Different \(\mathrm{h}_{m}\), same \(\mathrm{R}\) (4) different \(\mathrm{h}_{-}\)and \(\mathrm{R}\)

Ram and Shyam are walking on two perpendicular tracks with speed \(3 \mathrm{~ms}^{-1}\) and \(4 \mathrm{~m} \mathrm{~s}^{-1}\), respectively. At a certain moment (say \(\mathrm{t}=0 \mathrm{~s}\) ), Ram and Shyam are at \(20 \mathrm{~m}\) and \(40 \mathrm{~m}\) away from the intersection of tracks, respectively, and moving towards the intersection of the tracks. During the motion the magnitude of velocity of ram with respect to Shyam is (1) \(1 \mathrm{~ms}^{1}\) (2) \(4 \mathrm{~ms}^{-1}\) (3) \(5 \mathrm{~ms}^{1}\) (4) \(7 \mathrm{~ms}^{-1}\)

Points \(A\) and \(C\) are on the horizontal ground and \(A\) and \(B\) are in same vertical plane at a distance of \(1500 \mathrm{~m}\). Simultaneously bullets are fired from \(A, B\) and \(C\) and they collide at \(O\). The bullet at \(B\) is fired at an angle of \(30^{\circ}\) with horizontal towards the ground at velocity \(100 \mathrm{~m} / \mathrm{s}\). The bullet at \(C\) is projected vertically upward at velocity of \(100 \mathrm{~m} / \mathrm{s}\). The bullet projected from \(A\) reaches its maximum height at \(O\). Find the elevation angle \(\angle \theta=\angle O A C\) : (1) \(60^{\circ}\) (2) \(45^{\circ}\) (3) \(30^{\circ}\) (4) \(15^{\circ}\)

A student throws soft balls out of the window at different angles to the horizontal. All soft balls have the same inity speed \(v=10 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\). It turns out that all soft balls' landing velocities make angles \(30^{\circ}\) or greater with the horizontal Find the height \(h\) (in \(\mathrm{m}\) ) of the window above the ground.
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Measurements

Read Explanation

Medical Physics

Read Explanation

Nuclear Physics

Read Explanation

Translational Dynamics

Read Explanation

Fluids

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.