Question

How can you show that the equation for impulse: Ft=mv_mu is dimensionally correct, where F=ma promptly U= initial velocity, v=final velocity, t=time?

Answer 1

To show that the equation for impulse, Ft = m(v - u), is dimensionally correct, we need to analyze the dimensions of each term in the equation.

Let's break down the dimensions of each term:

F: Force (dimension: [M][L][T]^-2) t: Time (dimension: [T]) m: Mass (dimension: [M]) v: Final Velocity (dimension: [L][T]^-1) u: Initial Velocity (dimension: [L][T]^-1)

Now, let's substitute the dimensions into the equation and see if they are consistent:

Left-hand side (LHS): Ft = [M][L][T]^-2 [T] = [M][L][T]^-1

Right-hand side (RHS): m(v - u) = [M]([L][T]^-1 - [L][T]^-1) = [M][L][T]^-1

Both the LHS and RHS have the same dimensions of [M][L][T]^-1, which means they are consistent.

Therefore, the equation Ft = m(v - u) is dimensionally correct.