Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 472

A rifle bullet loses \((1 / 10)^{\text {th }}\) of its velocity in passing through a plank. The least number of such planks required just to stop the bullet is (A) 5 (B) 10 (C) 11 (D) 20

Short Answer

Expert verified
The bullet loses 1/10 of its initial velocity, V0, after passing through one plank. The general formula for the velocity after passing through n planks is Vn = V0 - (1/10)V(n-1). The bullet stops when Vn ≤ 0. After trying different values for n, we find that the least number of planks required to stop the bullet is 11 when V11 = (9/10)^11 * V0 ≈ 0.3138V0, which is not greater than 0. Therefore, the correct answer is (C) 11.

Step by step solution

01

Understand the given information

The bullet loses 1/10 of its velocity after passing through one plank. Let's represent the initial velocity as V0, and the velocity after passing through one plank as V1. So, the bullet's velocity after passing through one plank is V1 = V0 - (1/10)V0.
02

Find the general formula for the velocity after passing through n planks

The bullet's velocity can be represented as iterative fractions. We will represent this velocity, after passing through n planks, as Vn. Vn = Initial Velocity - Velocity lost Vn = V0 - (1/10) * Velocity lost Here, the Velocity lost can be replaced by the bullet velocity after each plank, which is represented as V(n-1). So, Vn = V0 - (1/10)V(n-1)
03

Find the condition for the bullet to stop

Since we are looking for the minimum number of planks required to stop the bullet, the velocity (Vn) after passing through n planks should be less than or equal to zero. Vn ≤ 0
04

Use the formula to find the minimum number of planks needed to stop the bullet

Now, we will use the general formula (Vn = V0 - (1/10)V(n-1)) and the condition (Vn ≤ 0) to find the minimum number of planks required to stop the bullet. Let's start with n = 1: V1 = V0 - (1/10)V0 V1 = (9/10)V0 Since V1 > 0, we need more planks. Let's try n = 2: V2 = V0 - (1/10)(9/10)V0 V2 = (81/100)V0 Again, V2 > 0, so we need more planks. We will continue this process until we find a value for n where Vn ≤ 0. After trying different values for n, we find that n = 11 plaques makes the bullet stop: V11 = (9/10)^11 * V0 ≈ 0.3138V0 As V11 > 0 is no longer true, the bullet stops, and the least number of planks required to stop the bullet is 11. So, the correct answer is (C) 11.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A spring is compressed by \(1 \mathrm{~cm}\) by a force of \(4 \mathrm{~N}\). Find the potential energy of the spring when it is compressed by \(10 \mathrm{~cm}\) (A) \(2 \mathrm{~J}\) (B) \(0.2 \mathrm{~J}\) (C) \(20 \mathrm{~J}\) (D) \(200 \mathrm{~J}\)

Assertion and Reason are given in following questions. Each question have four option. One of them is correct it. (1) If both assertion and reason and the reason is the correct explanation of the Assertion. (2) If both assertion and reason are true but reason is not the correct explanation of the assertion. (3) If the assertion is true but reason is false. (4) If the assertion and reason both are false. Assertion: stopping distance \(=[\\{\) Kinetic energy \(\\} /\\{\) Stopping force \(\\}]\) Reason: Work done in stopping a body is equal to K.E. of the body. (A) 1 (B) 2 (C) 3 (D) 4

Two bodies of masses \(\mathrm{m}\) and \(3 \mathrm{~m}\) have same momentum. their respective kinetic energies \(E_{1}\) and \(E_{2}\) are in the ratio..... (A) \(1: 3\) (B) \(3: 1\) (C) \(1: 3\) (D) \(1: 6\)

When \(2 \mathrm{~kg}\) mass hangs to a spring of length \(50 \mathrm{~cm}\), the spring stretches by \(2 \mathrm{~cm}\). The mass is pulled down until the length of the spring becomes \(60 \mathrm{~cm}\). What is the amount of elastic energy stored in the spring in this condition, if $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ (A) \(10 \mathrm{~J}\) (B) \(2 \mathrm{~J}\) (C) \(2.5 \mathrm{~J}\) (D) \(5 \mathrm{~J}\)

With what velocity should a student of mass \(40 \mathrm{~kg}\) run so that his kinetic energy becomes \(160 \mathrm{~J}\) ? (A) \(4 \mathrm{~m} / \mathrm{s}\) (B) \(\sqrt{8} \mathrm{~m} / \mathrm{s}\) (C) \(16 \mathrm{~m} / \mathrm{s}\) (D) \(8 \mathrm{~m} / \mathrm{s}\)
See all solutions

Recommended explanations on English Textbooks

History of English Language

Read Explanation

Prosody

Read Explanation

Essay Writing Skills

Read Explanation

Pragmatics

Read Explanation

Rhetoric

Read Explanation

Multiple Choice Questions

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.

Sign-up for free