/*! 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 6 A body of mass \(m\) is moving w... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A body of mass \(m\) is moving with speed \(v\) when a constant force \(F\) newtons is applied to it in the direction of motion for \(t\) seconds: (a) The impulse of the force is \(F t\) newton second. (b) \(F t=m v\), (c) The body loses an amount of kinetic energy equal to \(F t\). (d) The final speed of the body is \(v+\frac{F t}{m}\).

Short Answer

Expert verified
Statements (a), (b), and (d) are correct; statement (c) is incorrect.

Step by step solution

01

Understanding Impulse

Impulse is given by the product of the force and the time for which it is applied. Therefore, the impulse of the force is indeed: \[ I = F \times t \] This confirms statement (a) is correct.
02

Relating Impulse to Momentum Change

Impulse can also be related to the change in momentum of the body. If a body of mass \( m \) is subjected to a force \( F \) for a time \( t \), the impulse \( F \times t \) equals the change in momentum: \[ F \times t = \text{change in momentum} \]
03

Initial and Final Momentum

Since the force is applied in the direction of motion, the initial momentum of the body is \( m \times v \). The final momentum after time \( t \) is \( m \times v_{f} \) where \( v_f \) is the final speed.
04

Calculating the Change in Momentum

The change in momentum is: \[ \text{change in momentum} = m \times v_{f} - m \times v \] From the impulse-momentum theorem we get: \[ F \times t = m \times v_{f} - m \times v \]
05

Finding the Final Speed

Rearranging the equation: \[ F \times t = m \times (v_{f} - v) \] Therefore, \[ v_{f} = v + \frac{ F \times t }{ m } \] This confirms statement (d) is correct.
06

Kinetic Energy Consideration

The initial kinetic energy is \( \frac{1}{2} m v^2 \). After the force is applied, the final kinetic energy is \( \frac{1}{2} m (v + \frac{ F \times t }{ m })^2 \). The change in kinetic energy is given by: \[ \Delta KE = \frac{1}{2} m (v + \frac{ F \times t }{ m })^2 - \frac{1}{2} m v^2 \] Simplifying this is not straightforward and does not simply equal \( F \times t \), so statement (c) is incorrect.

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 91Ó°ÊÓ!

Key Concepts

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

Impulse
When we talk about 'Impulse' in physics, we're discussing the product of force and the time duration over which the force is applied. Mathematically, impulse is described as:
\[ I = F \times t \]. The impulse can be visualized as the push or pull experienced by an object over a certain period. For example, in our given exercise, a constant force is applied over a specific duration, and this generates an impulse. Impulse is significant because it relates directly to the change in momentum of an object. To sum up, impulse provides a handy way of looking at sudden forces acting over short times.
Momentum
Momentum, symbolized as \(p\), is the product of an object's mass \(m\) and its velocity \(v\). Essentially, momentum measures how hard it is to stop a moving object. If you consider our problem, the initial momentum of the body is represented by \(m \times v\). When a force acts on it for a time \(t\), it changes this momentum. This can be expressed using the impulse-momentum theorem, which states:
\[ F \times t = \text{change in momentum} \]. So, after time \(t\), the body's final momentum is \(m \times v_f\), where \(v_f\) is the new speed. By understanding momentum, it's easier to grasp how forces cause changes in the motion of an object.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is given by the equation:
\[ KE = \frac{1}{2} m v^2 \]. In the given exercise, the initial kinetic energy of the body before the force is applied is \(\frac{1}{2} m v^2\). After applying the force, the final kinetic energy becomes \(\frac{1}{2} m (v_f)^2\). The change in kinetic energy is more complex to compute directly, and it doesn't simply equate to the impulse (\(F \times t\)). This discrepancy shows kinetic energy's more intricate relationship with force and motion, emphasizing that energy transformations aren't always straightforward.
Force
Force is an interaction that changes the motion of an object. In our context, the constant force \(F\) is applied in the same direction as the object's motion. Using Newton's second law, we know force is related to mass and acceleration by \(F = m \times a\). When force is applied over time, it changes the object's momentum, calculated through impulse. Constant force means that the object continually accelerates while the force is applied. This results in a final velocity \(v_f\) which can be determined by the equation:
\[ v_f = v + \frac{F \times t}{m} \]. This clearly ties force to changes in motion, showing how sustained force affects an object's velocity and overall movement.

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

An inelastic string has a particle \(\mathrm{A}\) attached to one end and a particle \(\mathrm{B}\) attached to the other end. If \(\mathrm{A}\) is projected in the direction \(\overrightarrow{\mathrm{BA}}\) find the initial speed of \(\mathrm{B}\) if: (a) initially the string is slack. (b) the speed of projection of \(\mathrm{A}\) is \(4 \mathrm{~ms}^{-1}\), (c) the particles are of equal mass, (d) the string is \(2 \mathrm{~m}\) long.

State the law of conservation of linear momentum for two interacting particles. Show how the law of conservation of linear momentum applied to two particles which collide directly follows from Newton's laws of motion. Three smooth spheres A, B, C. equal in all respects, lie at rest and separated from one another on a smooth horizontal table in the order \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) with their centres in a straight line. Sphere A is projected with speed \(V\) directly towards sphere \(\mathrm{B}\). If the coefficient of restitution at each collision is \(e\), where \(0

Two particles \(\mathrm{A}\) and \(\mathrm{B}\) collide directily head-on and bounce. Find their speeds immediately after impact. (a) The mass of \(\mathrm{A}\) is twice the mass of B. (b) Just before impact the speed of \(\mathrm{A}\) is \(4 \mathrm{~ms}^{-1}\) and that of \(\mathrm{B}\) is \(3 \mathrm{~ms}^{-1} .\) (c) No kinetic energy is lost by the impact.

(a) The coefficient of restitution between two colliding objects is less than } 1 \text {. }

A smooth plane is fixed at an inclination \(30^{\circ}\) with its lower edge at a height \(a\) above a horizontal table. Two particles \(\mathrm{P}\) and \(\mathrm{Q}\), each of mass \(m\), are connected by a light inextensible string of length \(2 a\), and \(\mathrm{P}\) is held at the lower edge of the inclined plane while \(\mathbf{Q}\) rests on the table vertically below \(\mathrm{P}\). The particle \(\mathrm{P}\) is then projected with velocity \(u(u>\sqrt{g a}\) ) upwards along a line of greatest slope of the plane. Find the impulsive tension in the string when \(\mathrm{Q}\) is jerked into motion. Determine the magnitude of \(u\) if \(\mathrm{Q}\) just reaches the lower edge of the plane, and the tension in the string while \(Q\) is moving.

See all solutions

Recommended explanations on Math Textbooks

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.