Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 353

A \(0.5 \mathrm{~kg}\) ball moving with a speed of \(12 \mathrm{~ms}^{-1}\) strikes a hard wall at an angle of \(30^{\circ}\) with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for \(0.25 \mathrm{~S}\) the average force acting on the wall is (A) \(96 \mathrm{~N}\) (B) \(48 \mathrm{~N}\) (C) \(24 \mathrm{~N}\) (D) \(12 \mathrm{~N}\)

Short Answer

Expert verified
The average force acting on the wall is \(48 \mathrm{~N}\), which corresponds to option (B).

Step by step solution

01

Calculate the initial horizontal and vertical components of the momentum

First, we need to calculate the initial momentum components of the ball. The vertical component remains unchanged, while the horizontal component changes its direction after the collision. To calculate the components of the momentum, we use the mass of the ball and its initial speed, along with the given angle. Initial horizontal momentum: \(p_{x1} = mv\cos(\theta)\) Initial vertical momentum: \(p_{y1} = mv\sin(\theta)\)
02

Calculate the final horizontal and vertical components of the momentum

After the collision, the ball maintains its speed, and therefore, the vertical component of the momentum remains unchanged, while the horizontal component changes direction. We can calculate the final components of the momentum using the same formulas as before. Final horizontal momentum: \(p_{x2} = -mv\cos(\theta)\) (negative sign because the direction is opposite to the initial direction) Final vertical momentum: \(p_{y2} = mv\sin(\theta)\)
03

Calculate the change in the momentum of the ball

To calculate the change in the momentum, we will subtract the initial momentum components from the final momentum components. Change in horizontal momentum: \(\Delta p_x = p_{x2} - p_{x1} = -mv\cos(\theta) - mv\cos(\theta) = -2mv\cos(\theta)\) Change in vertical momentum: \(\Delta p_y = p_{y2} - p_{y1} = mv\sin(\theta) - mv\sin(\theta) = 0\)
04

Calculate the average force exerted on the wall

Now, we will calculate the average force exerted on the wall by the ball. For this, we will use the change in horizontal momentum and the time for which the ball is in contact with the wall. Average force: \(F_{avg} = \frac{\Delta p_x}{\Delta t}\)
05

Plug in the given values and solve for the average force

Now that we have the formula for the average force, let's plug in the given values and solve for it. \(F_{avg} = \frac{-2(0.5 \mathrm{~kg})(12 \mathrm{~ms}^{-1})\cos(30^{\circ})}{0.25 \mathrm{~S}}\) Calculating this expression, we get: \(F_{avg} = 48 \mathrm{~N}\) The average force acting on the wall is 48 N, which corresponds to option (B).

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 block B, placed on a horizontal surface is pulled with initial velocity \(\mathrm{V}\). If the coefficient of kinetic friction between surface and block is \(\mu\), than after how much time, block will come to rest? (A) (v/g) (B) \((\mathrm{g} / \mathrm{v})\) (C) \((\mathrm{g} / \mathrm{v})\) (D) \((\mathrm{v} / \mathrm{g})\)

Three Forces \(F_{1}, F_{2}\), and \(F_{3}\) together keep a body in equilibrium. If \(F_{1}=3 \mathrm{~N}\) along the positive \(\mathrm{X}\) - axis, \(\mathrm{F}_{2}=4 \mathrm{~N}\) along the positive Y-axis then the third force \(F_{3}\) is (A) \(5 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(3 / 4)\) with negative \(\mathrm{y}\) -axis (B) \(5 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(4 / 3)\) with negative \(\mathrm{y}\) -axis (C) \(7 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(3 / 4)\) with negative \(\mathrm{y}\) -axis (D) \(7 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(4 / 3)\) with negative \(\mathrm{y}\) -axis

An impulsive force of \(100 \mathrm{~N}\) acts on a body for \(1 \mathrm{sec}\) What is the change in its linear momentum ? (A) \(10 \mathrm{~N}-\mathrm{S}\) (B) \(100 \mathrm{~N}-\mathrm{S}\) (C) \(1000 \mathrm{~N}-\mathrm{S}\) (D) \(1 \mathrm{~N}-\mathrm{S}\)

10,000 small balls, each weighing \(1 \mathrm{~g}\) strike one square \(\mathrm{cm}\) of area per second with a velocity \(100 \mathrm{~ms}^{-1}\) in a normal direction and rebound with the same velocity. The value of pressure on the surface will be (A) \(2 \times 10^{3} \mathrm{Nm}^{-2}\) (B) \(2 \times 10^{5} \mathrm{Nm}^{-2}\) (C) \(10^{7} \mathrm{Nm}^{-2}\) (D) \(2 \times 10^{7} \mathrm{Nm}^{2}\)

The upper half of an inclined plane of inclination \(\theta\) is perfectly smooth while the lower half is rough A body starting from the rest at top come back to rest at the bottom, then the coefficient of friction for the lower half is given by (A) \(\mu=\sin \theta\) (B) \(\mu=\cot \theta\) (C) \(\mu=2 \cos \theta\) (D) \(\mu=2 \tan \theta\)
See all solutions

Recommended explanations on English Textbooks

English Language Study

Read Explanation

Textual Analysis

Read Explanation

Synthesis Essay

Read Explanation

Morphology

Read Explanation

English Grammar

Read Explanation

Global English

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