Identifying Dimensions Identical to Linear Momentum

This solution explores the concept of dimensional analysis in physics to determine which physical quantity shares the same dimensions as linear momentum. Dimensional analysis is a powerful tool that helps us understand the relationships between different physical quantities by examining their fundamental units of measurement (like Mass [M], Length [L], and Time [T]).

Understanding the Dimensions of Linear Momentum

First, let's establish the dimensions of linear momentum. Linear momentum ($p$) is defined as the product of an object's mass ($m$) and its velocity ($v$).

The formula for linear momentum is:

$ p = m \times v $

Now, let's represent this in terms of fundamental dimensions:

• The dimension of mass, $[m]$, is represented as [M].
• The dimension of velocity, $[v]$, which is distance over time, is represented as [L T^-1].

Therefore, the dimensions of linear momentum $[p]$ are obtained by multiplying the dimensions of mass and velocity:

$ [p] = [m] \times [v] = [M] \times [L T^{-1}] = [M L T^{-1}] $

Our task is to find which of the given options results in these same dimensions: [M L T^-1].

Analyzing the Dimensions of Each Option

We will now calculate the dimensions for each of the provided expressions:

Option 1 Analysis: Angular momentum divided by mass

Let's first find the dimensions of angular momentum ($L$). It's typically calculated as the product of moment of inertia ($I$) and angular velocity ($\omega$). The moment of inertia for a point mass is $I = m r^2$.

Dimensions breakdown:

• Dimension of moment of inertia $[I]$ = $[m] \times [r^2] = [M L^2]$
• Dimension of angular velocity $[\omega]$ = $[T^{-1}]$ (since angle is dimensionless)
• Therefore, the dimension of angular momentum $[L]$ = $[I] \times [\omega] = [M L^2 T^{-1}]$
• The dimension of mass $[m]$ is $[M]$.

Now, we calculate the dimensions of 'Angular momentum divided by mass':

$ \frac{[L]}{[m]} = \frac{[M L^2 T^{-1}]}{[M]} = [L^2 T^{-1}] $

Comparing this result, [L² T^-1], with the dimensions of linear momentum, [M L T^-1], we see they are not identical.

Option 2 Analysis: Force divided by time

We know that force ($F$) is mass times acceleration ($a$).

Dimensions breakdown:

• Dimension of force $[F]$ = $[m] \times [a] = [M] \times [L T^{-2}] = [M L T^{-2}]$
• The dimension of time $[T]$ is simply $[T]$.

Now, let's find the dimensions of 'Force divided by time':

$ \frac{[F]}{[T]} = \frac{[M L T^{-2}]}{[T]} = [M L T^{-3}] $

These dimensions, [M L T^-3], do not match the dimensions of linear momentum, [M L T^-1].

Option 3 Analysis: Power multiplied by length

Power ($P$) is the rate of doing work, or energy transfer per unit time. Work ($W$) is force applied over a distance.

Dimensions breakdown:

• Dimension of work $[W]$ = $[F] \times [distance] = [M L T^{-2}] \times [L] = [M L^2 T^{-2}]$
• Dimension of power $[P]$ = $\frac{[W]}{[T]} = \frac{[M L^2 T^{-2}]}{[T]} = [M L^2 T^{-3}]$
• The dimension of length $[L]$ is $[L]$.

Calculating the dimensions of 'Power multiplied by length':

$ [P] \times [L] = [M L^2 T^{-3}] \times [L] = [M L^3 T^{-3}] $

These dimensions, [M L³ T^-3], are different from the dimensions of linear momentum, [M L T^-1].

Option 4 Analysis: Energy divided by velocity

Let's consider the dimensions of energy ($E$). Kinetic energy, for example, is given by $E = \frac{1}{2}mv^2$.

Dimensions breakdown:

• Dimension of energy $[E]$ = $[m] \times [v^2] = [M] \times [L T^{-1}]^2 = [M L^2 T^{-2}]$
• The dimension of velocity $[v]$ is $[L T^{-1}]$.

Now, we calculate the dimensions of 'Energy divided by velocity':

$ \frac{[E]}{[v]} = \frac{[M L^2 T^{-2}]}{[L T^{-1}]} $

Simplifying the exponents:

$ [M L^{(2-1)} T^{(-2 - (-1))}] = [M L^1 T^{-1}] = [M L T^{-1}] $

These dimensions, [M L T^-1], are identical to the dimensions of linear momentum.

Final Conclusion

Through careful dimensional analysis, we have found that the expression 'Energy divided by velocity' yields the dimensions [M L T^-1], which is precisely the same as the dimensions of linear momentum.