Help with dimensional analysis (Phys first year)?

Hello! I have a question in my text book asking me to perform dimensional analysis on each one asking which are correct. I know the answer is that all three are correct but I don't understand how to do it. When I tried I must have been doing something wrong because I can't get any of them to work out.

1) 1/2mv^2 = 1/2mv(initial) + sqrt(mgh)

2) v = v(initial) + at^2

3) ma = v^2

2 Answers

Relevance
  • 8 years ago
    Favorite Answer

    Dimensional analysis is used to test whether an equation is correct. If it is correct, the following two rules apply:

    1. In order to add or subtract quantities, they should have same dimensions.

    2. Dimensions of the left hand side of the equation should be equal to the right hand side.

    Typically, we use M for mass, L for length or distance and T for time.

    As such, the dimension of V = distance/time = L/T=LT^-1

    Similarly, a = g =LT^-2

    Let's evaluate your questions:

    1. Check whether it satisfies the first rule:

    1/2mv = MLT^-1 (Note: constants don't have dimensions)

    sqrt(mgh) = sqrt (M*LT^-2*L) = ML^2T^-2

    Since the added ones don't have same dimensions, this equation is not correct.

    2. Similar procedure,

    v = LT^-1 and at^2= LT^-2*T^2 = L, again not satisfied the first rule. So wrong

    3. LHS = ma = MLT^-2

    RHS = V^2 = *(LT^-2)^2 = L^2T^-4

    Not satisfied. So, wrong

    Hope it helps

    • Commenter avatarLogin to reply the answers
  • 4 years ago

    In elemantry, all of my phys coaching instructors have been female slightly on the strict area, with ponytails. For some unusual reason absolutely everyone enjoyed them. Now I certainly have a guy who's relating to the comparable height as my dad, 5 foot 4 with blond curly hair and glasses.... the only words that pop out of his mouth are "stable cases." Lord its' particularly annoying.

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.