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Chapter 6: Problem 21

If the momentum of a body increases by \(0.01 \%\), its kinetic energy will increase by (a) \(0.01 \%\) (b) \(0.02 \%\) (c) \(0.04 \%\) (d) \(0.08 \%\)

Short Answer

Expert verified
The kinetic energy will increase by (b) 0.02 %.

Step by step solution

01

Understand the relationship.

First, observe that kinetic energy (KE) is proportional to the square of velocity (v^2), while momentum (P) is proportional to velocity (v). This means that a percentage change in momentum translates to a different percentage change in kinetic energy. When momentum increases by a certain percentage, the velocity, and hence kinetic energy, will also increase.
02

Calculate the percentage increase in kinetic energy.

Have a 0.01% increase in v due to the increase in momentum. Because KE is proportional to the square of velocity, the change in KE will be given by the square of the change in velocity. Therefore, the increase in KE will be (1 + 0.01/100)^2 - 1, expressed as a percentage.
03

Simplify and express as percentage.

Simplify the expression to find the percentage increase. Thus, it comes out to be approximately 0.02%.

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Key Concepts

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

Momentum
Momentum is a measure of the motion of an object and is calculated as the product of its mass and velocity. It is an important concept in physics because it reflects the degree of motion and is conserved in isolated systems.

How to calculate momentum:
  • The formula to calculate momentum (P) is: \( P = m \times v \), where \( m \) is the mass and \( v \) is the velocity of the object.
  • Understanding that momentum is proportional to velocity helps in predicting how changes in velocity could affect other quantities like kinetic energy.
An increase in momentum indicates an increase in velocity if mass remains constant. This is crucial for solving problems involving changes in kinetic energy, as seen in the exercise above. A small increase in momentum, such as 0.01%, results in a proportional increase in velocity and subsequently affects kinetic energy.
Velocity
Velocity is the speed of an object in a specified direction. It is a vector quantity, which means it has both magnitude and direction, making it distinct from speed alone. Velocity plays a crucial role in understanding motion and dynamics in physics.

Key points about velocity:
  • Velocity determines how quickly an object's position changes over time in a certain direction.
  • In the exercise, a change in momentum implies a change in velocity, which affects kinetic energy.
Since kinetic energy is proportional to the square of velocity, any change in velocity results in a squared change in kinetic energy. For example, a tiny increase in momentum leads to a slight increase in velocity, but the effect on kinetic energy is more pronounced because of the squaring effect.
Percentage Increase
Percentage increase is a way to express how much a quantity has changed relative to its original value. It鈥檚 a common method used to understand the relative growth of a value in a simple numerical format.

How to calculate percentage increase:
  • The formula is: \(\text{Percentage Increase} = \left( \frac{\text{New Value} - \text{Old Value}}{\text{Old Value}} \right) \times 100\% \)
  • It鈥檚 useful for comparing changes in variables that may initially seem small, like the 0.01% increase in momentum in the exercise.
In the exercise context, we see a momentum increase of 0.01%, which is a very small change. Due to the squaring relationship between velocity and kinetic energy, this small percentage increase in momentum results in approximately a 0.02% increase in kinetic energy. This highlights the importance of understanding how percentage changes in one physical quantity can lead to significant changes in others.

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Most popular questions from this chapter

A block of mass \(2 \mathrm{~kg}\) is released from \(A\) on the track that is one quadrant of a circle of radius \(1 \mathrm{~m}\). It slides down the track and reaches \(B\) with a speed of \(4 \mathrm{~ms}^{-1}\) and finally stops at \(C\) at a distance of \(3 m\) from \(B\). The work done against the force of friction is (a) \(10 J\) (b) \(20 J\) (c) \(2 J\) (d) \(6 J\)

A sphere of mass \(0.1 \mathrm{~kg}\) is attached to a cord of \(1 \mathrm{~m}\) length. Starting from the height of its point of suspension this sphere hits a block of same mass at rest on a frictionless table, If the impact is elastic, then the kinetic energy of the block after the collisio (a) \(1 J\) (b) \(10 J\) (c) \(0.1 J\) (d) \(0.5 \mathrm{~J}\)

A bus weighing 100 quintals moves on a rough road with a constant speed of \(72 \mathrm{~km} / \mathrm{h}\). The friction of the road is \(9 \%\) of its weight and that of air is \(1 \%\) of its weight. What is the power of the engine. Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) (a) \(50 \mathrm{~kW}\) (b) \(100 \mathrm{~kW}\) (c) \(150 \mathrm{~kW}\) (d) \(200 \mathrm{~kW}\)

The vessels \(A\) and \(B\) of equal volume and weight are immersed in water to a depth \(h\). The vessel \(A\) has an opening at the bottom through which water can enter. If the work done in immersing \(A\) and \(B\) are \(W_{A}\) and \(W_{B}\) respectively, then (a) \(W_{A}=W_{B}\) (b) \(W_{A}W_{B}\) (d) \(W_{A}>=\left\langle W_{B}\right.\)

A pump motor is used to deliver water at a certain rate from a given pipe. To obtain twice as much water from the same pipe in the same time, power of the motor has to be increased to (a) 16 times (b) 4 times (c) 8 times (d) 2 times
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