If an object of mass 10 kg is moving with a uniform speed of 10 m/s, then the linear momentum and the kinetic energy of the object, respectively, are
100 N.s and 500 J
This problem asks us to find the linear momentum and kinetic energy of an object given its mass and speed. We need to apply the standard formulas for these two physical quantities.
From the question, we are given:
Since the object is moving with a uniform speed in a specific direction, we can consider the magnitude of the velocity to be equal to the speed for calculating momentum.
We use the formula for linear momentum:
\(p = m \times v\)
Substitute the given values:
\(p = 10 \, \text{kg} \times 10 \, \text{m/s}\)
Calculate the product:
\(p = 100 \, \text{kg} \cdot \text{m/s}\)
The unit kg⋅m/s is equivalent to N⋅s. So, the linear momentum is \(100 \, \text{N} \cdot \text{s}\).
We use the formula for kinetic energy:
\(KE = \frac{1}{2} m v^2\)
Substitute the given values:
\(KE = \frac{1}{2} \times 10 \, \text{kg} \times (10 \, \text{m/s})^2\)
First, calculate the square of the speed:
\(v^2 = (10 \, \text{m/s})^2 = 100 \, \text{m}^2/\text{s}^2\)
Now substitute this back into the KE formula:
\(KE = \frac{1}{2} \times 10 \, \text{kg} \times 100 \, \text{m}^2/\text{s}^2\)
Calculate the product:
\(KE = 5 \, \text{kg} \times 100 \, \text{m}^2/\text{s}^2\)
\(KE = 500 \, \text{kg} \cdot \text{m}^2/\text{s}^2\)
The unit kg⋅m²/s² is equivalent to Joules (J). So, the kinetic energy is \(500 \, \text{J}\).
The question asks for the linear momentum and kinetic energy respectively. Therefore, the answer should be presented as \(100 \, \text{N} \cdot \text{s}\) and \(500 \, \text{J}\).
Let's compare our calculated values with the given options:
| Option | Linear Momentum | Kinetic Energy | Matches Calculation? |
|---|---|---|---|
| 1 | 100 N.s | 500 J | Yes |
| 2 | 100 N.s | 1000 J | No (KE is wrong) |
| 3 | 200 N.s | 500 J | No (Momentum is wrong) |
| 4 | 200 N.s | 1000 J | No (Both are wrong) |
Option 1 correctly matches our calculated values for both linear momentum and kinetic energy.
| Concept | Formula | Units |
|---|---|---|
| Linear Momentum (\(p\)) | \(p = m v\) | kg⋅m/s or N⋅s |
| Kinetic Energy (KE) | \(KE = \frac{1}{2} m v^2\) | kg⋅m²/s² or J (Joules) |
| Mass (\(m\)) | - | kg (kilograms) |
| Velocity (\(v\)) / Speed (\(v\)) | - | m/s (meters per second) |
Linear momentum is a vector quantity, meaning it has both magnitude and direction. In this problem, we were given speed, which is the magnitude of velocity. For kinetic energy, which is a scalar quantity, we only need the magnitude of velocity (speed).
The concept of momentum is related to Newton's laws of motion, particularly the second law (\(F_{net} = \frac{\Delta p}{\Delta t}\)). Energy, including kinetic energy, is a scalar quantity that represents the capacity to do work. The work-energy theorem relates the net work done on an object to its change in kinetic energy.
The following diagram shows a pendulum at different positions. Which one of the following statement is true?

Which one of the following does not convert electrical energy into light energy?
A microphone converts:
Which of the following is also called the First Law of Thermodynamics?
Select the most appropriate option to fill in the blanks. A thermistor can have a region of _______ determined by the construction material or the temperature of the material. The change in temperature can be due to internal effects such as current through the thermistor or due to external effects of _____.
EMF of a thermocouple is approximately a _____ function of the temperature difference between the junctions.
A particle of mass m starts from rest and moves with a constant acceleration a along a straight line. What is the relationship between the distance travelled x and the kinetic energy K of the particle?