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If a force of \(250 \mathrm{~N}\) act on body the momentum acquired is \(125 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\). What is the period for which force acts on the body? (a) \(0.5 \mathrm{~s}\) (b) \(0.2 \mathrm{~s}\) (c) \(0.4 \mathrm{~s}\) (d) \(0.25 \mathrm{~s}\)

Short Answer

Expert verified
The force acts on the body for 0.5 seconds.

Step by step solution

01

Understanding the Principle

The problem involves a force acting on a body and the momentum acquired by the body. We can relate force, time, and momentum using the formula: Force \( (F) \times \text{time} (t) = \text{Change in momentum} (\Delta p) \). This is derived from Newton's second law of motion.
02

Identify Given Values

From the problem, we know the force \( F = 250 \text{ N} \) and the acquired momentum (change in momentum) \( \Delta p = 125 \text{ kg} \cdot \text{m/s} \). We need to find the time \( t \).
03

Rearrange the Formula for Time

To find the time \( t \), rearrange the formula to solve for \( t \): \[ t = \frac{\Delta p}{F} \].
04

Substitute Values and Solve

Substitute the known values into the equation: \[ t = \frac{125 \text{ kg} \cdot \text{m/s}}{250 \text{ N}} \]. Simplify the fraction: \[ t = \frac{125}{250} = 0.5 \text{ s} \].

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Key Concepts

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

Force and Momentum Relation
Newton's Second Law forms the foundation of understanding the relationship between force and momentum. The law broadly states that the force acting on an object is equal to the rate of change of its momentum. Mathematically, this can be expressed as:
\[ F = \frac{\Delta p}{\Delta t} \]This formula conveys that force \( F \) is directly linked to the change in momentum \( \Delta p \) over time \( \Delta t \). It's a powerful tool for analyzing how forces impact motion, making the analysis of events involving acceleration or deceleration much clearer. Through this relationship:
  • Greater force results in a more considerable change in momentum over a given time.
  • Similarly, shorter force application times will result in a smaller momentum change if the force is constant.
Understanding this relation helps bridge concepts between force (a tangible influence capable of causing motion) and momentum (a property measuring motion's resistance to change).
Impulse
Impulse is a concept deeply tied to the force and momentum relation. It is defined as the change in momentum of an object when a force is applied over a specific period. Impulse \( J \) can be calculated using:
\[ J = F \times \Delta t = \Delta p \]Impulse appears in many physical situations, such as when catching a ball or applying brakes in a vehicle. The key points about impulse:
  • Impulse is vector quantity, meaning it has both a magnitude and a direction.
  • It describes the effect of a time-dependent force on an object.
  • It essentially "records" how much the momentum has changed due to the force applied.
In any real-world application, understanding impulse allows us to comprehend how forces, applied over time, cause the momentum of objects to change, influencing how objects move or stop.
Change in Momentum
The change in momentum is a central concept when analyzing motions and forces. When a force is applied to a body, it alters the momentum of the body, which is given by:
\[ \Delta p = m \cdot \Delta v \]where \( \Delta p \) is the change in momentum, \( m \) is mass, and \( \Delta v \) is the change in velocity.
For understanding the change in momentum:
  • Momentum changes can happen due to forces initiations, terminations, or variations.
  • Larger applied forces or longer durations of such forces generally cause larger momentum changes.
  • It is a vital consideration in collision analyses, where sudden momentum changes occur.
Applying these principles to our initial exercise, the change in momentum was specified as 125 kg-m/s, directly reflecting the impact of the 250 N force applied over a half-second duration. Understanding momentum changes illuminates numerous phenomena within the physical world, from simple pushes to complex machine operations.

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