/*! 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 96 A mass \(A\) is released from th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 96

A mass \(A\) is released from the top of a frictionless inclined plane \(18 \mathrm{~m}\) long and reaches the bottom 3 second later. At the instant when \(A\) is released, a second mass \(B\) is projected upwards along the plane from the bottom with a certain initial velocity. The mass \(B\) travels a distance up the plane, stops and returns to the bottom simultaneously with \(A\). The two masses do not collide. Initial velocity of \(B\) is (A) \(4 \mathrm{~ms}^{-1}\) (B) \(5 \mathrm{~ms}^{-1}\) (C) \(6 \mathrm{~ms}^{-1}\) (D) \(7 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The initial velocity of mass B is (C) \(6 \mathrm{~ms}^{-1}\).

Step by step solution

01

Calculate Acceleration of Mass A

: As mass A is released from the top of the frictionless inclined plane, it reaches the bottom in 3 seconds. We know the length of the plane is \(18 m\). Using the equation of motion, \(s = ut + \frac{1}{2}at^2 \), where, \(s\) is the distance covered by the mass A in 18 m, \(u\) is the initial velocity which is \(0\), \(a\) is the acceleration of the mass A, and \(t\) is the time taken which is \(3s\). In this case, \(s=18m\). So, the equation becomes \(18 = 0 \cdot 3 + \frac{1}{2}a (3)^2\).
02

Calculate Distance Travelled by Mass B

: Let \(x\) be the distance travelled by mass B up the incline before it stops. Since mass B stops and returns to the bottom at the same time as mass A, it will take mass B 3 seconds to travel the distance \(2x\) up and back down. Using the equation of motion, \(2x = ut + \frac{1}{2}at^2 \), where, \(2x\) is the total distance covered by mass B, \(u\) is the initial velocity that we need to find, \(a\) is the acceleration which is the same as that of mass A, and \( t \) is the total time taken which is \(3s\). Now we have to plug in the value of acceleration from step 1 and solve for the initial velocity \(u\).
03

Solve for the Initial Velocity of Mass B

: First, solve for acceleration from step 1: \(18 = \frac{1}{2}a (3)^2\) => \(a = \frac{18 \cdot 2}{3^2}\) => \(a = 4 \mathrm{~ms}^{-2}\) Now, plug in the value of acceleration into the equation of motion for mass B: \(2x = u \cdot 3 + \frac{1}{2}(4)(3)^2\) => \(2x = 3u + 18\) From the equation of motion for mass A, we can also write the relation between x and the acceleration: \(x = \frac{1}{2}at^2\) =>\(x = \frac{1}{2}(4)(3)^2\) => \(x = 18\) Now, plug in this value for x into the equation for mass B: \(2(18) = 3u + 18\) => \(36 = 3u + 18\) => \(u = \frac{36 - 18}{3}\) => \(u = 6 \mathrm{~ms}^{-1}\) Therefore, the correct answer is (C) \(6 \mathrm{~ms}^{-1}\).

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!

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

Statement-1: Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement-2: Principle of conservation of momentum holds true for all kinds of collisions. (A) Statement-1 is true, Statement- 2 is true; Statement- 2 is the correct explanation of Statement- 1 (B) Statement-1 is true, Statement- 2 is true; Statement- 2 is not the correct explanation of Statement-1 (C) Statement- 1 is false, Statement- 2 is true. (D) Statement- 1 is true, Statement- 2 is false.

A particle of mass \(m\) and velocity \(\vec{v}\) collides elastically with a stationary particle of same mass \(m\). If the collision is oblique, then the angle between the velocity vectors of the two particles after the collision is (A) \(\frac{\pi}{4}\) (B) \(\frac{\pi}{3}\) (C) \(\frac{\pi}{2}\) (D) \(\pi\)

Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. Then, if the spheres roll without slipping. (A) the heavier sphere reaches the bottom first. (B) the bigger sphere reaches the bottom first. (C) the two spheres reach the bottom together. (D) velocity of the both sphere will be same at the bottom.

A \(3 \mathrm{~kg}\) ball strikes a heavy rigid wall with a speed of \(10 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) with the wall. It gets reflected with the same speed at \(60^{\circ}\) with the wall. If the ball is in contact with the wall for \(0.2 \mathrm{~s}\), the average force exerted on the ball by the wall is (A) \(300 \mathrm{~N}\) (B) Zero (C) \(150 \sqrt{3} \mathrm{~N}\) (D) \(150 \mathrm{~N}\)

A body of mass \(m_{1}\) moving with uniform velocity of \(40 \mathrm{~m} / \mathrm{s}\) collides with another mass \(m_{2}\) at rest and then the two together begin to move with uniform velocity of \(30 \mathrm{~m} / \mathrm{s}\). The ratio of their masses \(\frac{m_{1}}{m_{2}}\) is (A) \(0.75\) (B) \(1.33\) (C) \(3.0\) (D) \(4.0\)
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Quantum Physics

Read Explanation

Electrostatics

Read Explanation

Physical Quantities And Units

Read Explanation

Nuclear Physics

Read Explanation

Torque and Rotational Motion

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.