/*! 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 2 A particle of mass \(2 \mathrm{~... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 2

A particle of mass \(2 \mathrm{~kg}\) moving with a velocity of \(3 \mathrm{~m} / \mathrm{s}\) is acted upon by a force which changes its direction of motion by an angle of \(90^{\circ}\) without changing its speed. What is the magnitude of impulse experienced by the particle? (1) \(6 \mathrm{~N} \mathrm{~s}\) (2) \(2 \mathrm{~N} \mathrm{~s}\) (3) \(3 \sqrt{2} \mathrm{~N} \mathrm{~s}\) (4) \(6 \sqrt{2} \mathrm{~N} \mathrm{~s}\)

Short Answer

Expert verified
The magnitude of the impulse is \(6\sqrt{2} \mathrm{~N~s}\) (option 4).

Step by step solution

01

Calculate initial momentum

The initial momentum of the particle is calculated using the formula \( p = mv \), where \( m = 2 \mathrm{~kg} \) and \( v = 3 \mathrm{~m/s} \). Thus, the initial momentum \( p_i = 2 \times 3 = 6 \mathrm{~kg~m/s} \).
02

Calculate final momentum

The speed remains the same, but the direction changes by \( 90^{\circ} \). Thus, the final momentum \( p_f \) also has a magnitude of \( 6 \mathrm{~kg~m/s} \), but is perpendicular to \( p_i \).
03

Determine the change in momentum

Since initial and final momenta are perpendicular, use the Pythagorean Theorem to find the change in momentum \( \Delta p \): \( \Delta p = \sqrt{p_i^2 + p_f^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \mathrm{~kg~m/s} \).
04

Relate impulse to change in momentum

Impulse \( J \) is equal to the change in momentum, \( J = \Delta p = 6\sqrt{2} \mathrm{~N~s} \). The magnitude of the impulse experienced by the particle is therefore \( 6\sqrt{2} \mathrm{~N~s} \).

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.

Change in Momentum
Momentum, in simple terms, is the product of an object's mass and its velocity. It's a vector quantity, meaning it has both magnitude and direction. In our scenario, when the force acts on a particle and changes its direction of motion by 90 degrees without altering its speed, the particle's momentum changes.
Change in momentum, often referred to as impulse, depends on how much the velocity changes in direction rather than magnitude.
  • The initial momentum: Before the force acts, the particle's momentum is directed along its initial path.
  • The final momentum: After the force acts, even though the speed remains the same, the momentum now points in a direction perpendicular to the initial momentum due to the 90-degree turn.
The change in momentum (94p) is the vector difference between these two perpendicular momenta, calculated using the Pythagorean Theorem, indicative of how the momentum's vector nature plays a crucial role in such situations.
Pythagorean Theorem in Physics
The Pythagorean Theorem, a concept from geometry, finds its relevance in physics, especially when dealing with vectors. In the context of our problem, since the initial and final momenta are vectors at a right angle to each other, the Pythagorean Theorem is perfectly suited to determine the change in magnitude of these vectors.
Here is how it applies:
  • The initial and final momenta magnitudes: Both are 6 kg m/s, forming the legs of a right triangle.
  • The change in momentum forms the hypotenuse, calculated as: \[\Delta p = \sqrt{p_i^2 + p_f^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \text{ kg m/s}.\]
The elegant transition from geometry to physics allows us to solve problems involving changes in velocity direction easily and effectively.
Force and Motion
In physics, force is what causes an object to move or change direction. Motion is the resulting change in an object's position due to applied force. This exercise showcases how force changes the direction rather than the speed of motion.
The momentary application of force:
  • The force changes the particle's momentum direction by 90 degrees.
  • This application of force leads to a change in momentum, evidenced by the impulse.
  • Even though the speed is constant, a directional change signifies a notable force impact, pivotal in reshaping the motion path, adding to this key concept.
Understanding force and motion helps explain diverse natural phenomena and the mechanics of daily activities.
Newton's Second Law
Newton's Second Law states that the force applied to an object equals the change in momentum per unit time (or the rate of momentum change), usually stated as \( F = \frac{dp}{dt} \). In simpler terms, the more force you apply, the greater the rate of velocity and, subsequently, momentum change.
This is intricately related to impulse, as impulse is the change in momentum resulting from such a force.
  • Impulse: This is the integral of force over the time it acts, directly equating to the change in momentum.
  • In the given problem, the force acts for an infinitesimally small period, causing a change in direction, underlining that force doesn鈥檛 need to act longer to produce a significant momentum change.
Newton鈥檚 Second Law gives strong support to understanding how forces affect motion, presenting a solid framework for analyzing dynamic systems and events in mechanics.

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

Three blocks are placed on smooth horizontal surface and lie on same horizontal straight line. Block 1 and block 3 have mass \(m\) each and block 2 has mass \(M(M>m) .\) Block 2 and block 3 are initially stationary, while block 1 is initially moving towards block 2 with speed \(v\) as shown. Assume that all collisions are head on and perfectly elastic. What value of \(M / m\) ensures that block \(\mid\) and block 3 have the same final speed? (1) \(5+\sqrt{2}\) (2) \(5-\sqrt{2}\) (3) \(2+\sqrt{5}\) (4) \(3+\sqrt{5}\)

A glass ball collides with a smooth horizontal surface ( \(x z\) plane) with a velocity \(V=a \hat{i}-b \hat{j} .\) If the coefficient of restitution of collision be \(e\), the velocity of the ball just afie the collision will be (1) \(\sqrt{e^{2} a^{2}+b^{2}}\) at angle \(\tan ^{-1}\left(\frac{a}{e b}\right)\) to the vertical (2) \(\sqrt{a^{2}+e^{2} b^{2}}\) at angle \(\tan ^{-1}\left(\frac{a}{e b}\right)\) to the vertical (3) \(\sqrt{a^{2}+\frac{b^{2}}{e^{2}}}\) at angle \(\tan ^{-1}\left(\frac{e a}{b}\right)\) to the vertical (4) \(\sqrt{\frac{a^{2}}{e^{2}}+b^{2}}\) at angle \(\tan ^{-1}\left(\frac{a}{e b}\right)\) to the vertical

Two spheres \(A\) and \(B\) move on a smooth horizontal surface with the same velocity \(V\) and have some separation between them. A third sphere \(C\) is moving in opposite direction on same surface with same speed. All the spheres are of equal? mass. The collisions are elastic. \(V_{\mathrm{CM}}\) represents the centre mass velocity of all the three spheres. Column II represnt the values after all the possible impacts have occurred. $$ \begin{array}{|l|l|} \hline {\text { Column I }} & {\text { Column II }} \\ \hline \begin{array}{l} \text { i. } \text { If } A \text { and } B \text { are not } \\ \text { connected to each other } \end{array} & \begin{array}{l} \text { a. } \\ \text { is } V / 3 \end{array} \\ \hline \begin{array}{l} \text { ii. If } A \text { and } B \text { are connected } \\ \text { to each other by a } \\ \text { massless rigid rod } \end{array} & \begin{array}{l} \text { b. } V_{\mathrm{CM}} \text { after all the } \\ \text { collision is } \frac{V}{3} \end{array} \\ \hline \begin{array}{l} \text { iii. If } A \text { and } B \text { are connected } \\ \text { by an ideal string } \end{array} & \begin{array}{r} \text { c. } \\ \text { bystem }(B+C) \text { is not } \\ \text { conserved } \end{array} \\ \hline \begin{array}{r} \text { iv. } \text { If } A \text { and } B \text { are connected } \\ \text { by an ideal spring which } \\ \text { is initially unstretched } \end{array} & \text { d. } \begin{array}{l} \text { Momentum of system } \\ (A+B) \text { is conserved } \end{array} \\ \hline \end{array} $$

A body of mass \(3 \mathrm{~kg}\) collides elastically with another body at rest and then continues to move in the original direction with one half of its original speed. What is the mass of the target body? (1) \(1 \mathrm{~kg}\) (2) \(1.5 \mathrm{~kg}\) (3) \(2 \mathrm{~kg}\) (4) \(5 \mathrm{~kg}\)

A block of mass \(m\) starts from rest and slides down a frictionless semi- circular track from a height \(h\) as shown. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass \(m\). If the block and the putty stick together and continue to slide, the maximum height that the block-putty system could reach is (1) \(h / 4\) (2) \(h / 2\) (3) \(h\) (4) independent of \(h\)
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Turning Points in Physics

Read Explanation

Modern Physics

Read Explanation

Medical Physics

Read Explanation

Oscillations

Read Explanation

Force

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.