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Chapter 6: Problem 13

While waiting in a car at a stoplight and \(80 \mathrm{~kg}\) man 19 and his car are suddenly accelerated to a speed of \(5 \mathrm{~ms}^{-1}\) as a result or rear end collision. If the time of impact is \(0.4 \mathrm{~s}\), find the average force on the man (a) \(100 \mathrm{~N}\) (b) \(200 \mathrm{~N}\) (c) \(500 \mathrm{~N}\) (d) \(1000 \mathrm{~N}\)

Short Answer

Expert verified
The average force on the man is 1000 N (option d).

Step by step solution

01

Understanding the Given Values

We are given that the man's mass is \( m = 80 \mathrm{~kg} \) and the change in speed (velocity) is \( \Delta v = 5 \mathrm{~m/s} \). The time of impact is \( t = 0.4 \mathrm{~s} \). We are asked to find the average force experienced by the man.
02

Use Newton's Second Law

To find the average force, we will use Newton's second law, which states that force is the rate of change of momentum. This can be mathematically expressed as: \( F = \frac{\Delta p}{\Delta t} \), where \( \Delta p \) is the change in momentum and \( \Delta t \) is the time over which the change occurs.
03

Calculate the Change in Momentum

The change in momentum \( \Delta p \) is the mass times the change in velocity: \( \Delta p = m \cdot \Delta v = 80 \mathrm{~kg} \cdot 5 \mathrm{~m/s} \). Thus, \( \Delta p = 400 \mathrm{~kg} \cdot \mathrm{m/s} \).
04

Compute the Average Force

Substitute the known values into the formula for force: \( F = \frac{\Delta p}{\Delta t} = \frac{400 \mathrm{~kg} \cdot \mathrm{m/s}}{0.4 \mathrm{~s}} \). This gives \( F = 1000 \mathrm{~N} \).
05

Conclusion

Comparing the calculated force with the options provided, the correct choice is (d) \(1000 \mathrm{~N}\).

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics that describes the motion of an object. It is defined as the product of an object's mass and its velocity. Mathematically, this can be expressed as:
  • Momentum (\( p \)) = Mass (\( m \)) \(\times\) Velocity (\( v \)).

In the context of this exercise, momentum changes when the car and its occupant are accelerated. The momentum change (\( \Delta p \)) is calculated as the mass of the man multiplied by the change in velocity (\( \Delta v \)):
  • \( \Delta p = m \times \Delta v \)
  • \( \Delta p = 80 \, \text{kg} \times 5 \, \text{m/s} \)
  • \( \Delta p = 400 \, \text{kg}\cdot\text{m/s} \)
This change in momentum reflects how much the object's state of motion has been altered during the collision. Understanding momentum is essential for grasping how forces affect motion over time.
Change in Velocity
The change in velocity is a crucial aspect when analyzing motion. It indicates how the speed of an object has changed in a certain timeframe, which is directly influenced by external forces. In this problem, the car and the man inside are initially at rest and suddenly accelerated to a velocity of \(5 \, \text{m/s} \).
The change in velocity (\( \Delta v \)) of the man as a result of the collision is calculated by subtracting the initial velocity from the final velocity:
  • Initial velocity \( v_i = 0 \, \text{m/s} \)
  • Final velocity \( v_f = 5 \, \text{m/s} \)
  • \( \Delta v = v_f - v_i = 5 \, \text{m/s} \)

This velocity change is significant as it determines the scale of momentum change, showcasing how much the object's motion has been modified. It's a building block for calculating the force exerted on the man during the collision.
Average Force
Average force is an important measure in physics that helps us understand the effect of forces over time. It is calculated as the total change in momentum divided by the time interval during which this change occurs:
  • \( F = \frac{\Delta p}{\Delta t} \)

In this exercise, the objective is to determine the force exerted on the man during the collision. Using the values:
  • \( \Delta p = 400 \, \text{kg}\cdot\text{m/s} \)
  • \( \Delta t = 0.4 \, \text{s} \)

The calculation of average force is:
  • \( F = \frac{400 \, \text{kg} \cdot \text{m/s}}{0.4 \, \text{s}} = 1000 \, \text{N} \)
This force (\(1000 \, \text{N}\)) represents the average push or pull exerted on the man due to the collision. Understanding how average force is derived gives insight into the dynamics of real-world scenarios, revealing how forces impact objects during interactions.

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Most popular questions from this chapter

A ball of mass \(0.2 \mathrm{~kg}\) is thrown vertically upwards by applying a force by hand. If the hand moves \(0.2 \mathrm{~m}\) while applying the force and the ball goes upto \(2 \mathrm{~m}\) height further, find the magnitude of the force [take \(\left.g=10 \mathrm{~ms}^{-2}\right]\) (a) \(16 \mathrm{~N}\) (b) \(20 \mathrm{~N}\) (c) \(22 \mathrm{~N}\) (d) \(4 \mathrm{~N}\)

80 railway wagons all of same mass \(5 \times 10^{3} \mathrm{~kg}\) are pulled by an engine with a force of \(4 \times 10^{5} \mathrm{~N}\). The tension in the coupling between \(30 t h\) and 31 st wagon from the engine is (a) \(25 \times 10^{4} \mathrm{~N}\) (b) \(40 \times 10^{4} \mathrm{~N}\) (c) \(20 \times 10^{4} \mathrm{~N}\) (d) \(32 \times 10^{4} \mathrm{~N}\)

\(100 \mathrm{~g}\) of an iron ball having velocity \(10 \mathrm{~ms}^{-1}\) collides with wall at an angle \(30^{\circ}\) and rebounds with the same angle. If the period of contact between the ball and wall is \(0.1 \mathrm{~s}\), then the average force experienced by the wall is (a) \(10 \mathrm{~N}\) (b) \(100 \mathrm{~N}\) (c) \(1.0 \mathrm{~N}\) (d) \(0.1 \mathrm{~N}\)

A shell is fired from a cannon with velocity \(v \mathrm{~ms}^{-1}\) at an angle \(\theta\) with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass, One of the pieces retraces its path to the cannon and the speed in \(\mathrm{m} / \mathrm{s}\) of the piece immediately after the explosion is (a) \(3 v \cos \theta\) (b) \(2 v \cos \theta\) (c) \(\frac{3 \mathrm{~V}}{2} \cos \theta\) (d) \(\frac{\sqrt{3} v \cos \theta}{2}\)

When forces \(F_{1}, F_{2}, F_{3}\) are acting on a particle of mass \(m\) such that \(F_{2}\) and \(F_{3}\) are mutually perpendicular, then the particle remains stationary. If the force \(F_{1}\) is now removed, then the acceleration of the particle is (a) \(F / m\) (b) \(E F_{1} m F_{i}\) \((c)\left(F_{2}-F\right) / m\) (d) \(F / m\)
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