Question

check the correctness of equation using dimensions analysis of v=u²+2as​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\textsf{ \implies Third equation of motion i.e. 2as = \sf v^2-u^2 }$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Check :- }}}$

$\textsf{ \implies The correctness of the equation dimensionally.}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

The equation given to us is 2as = v² - u². This is the third equation of motion which establishes the relationship between acclⁿ , displacement , final velocity and initial velocity . Here , a is acceleration , s is displacement , v is final velocity and u is initial velocity. So ,

$\sf\longrightarrow Dimensional \ formula \ of \ velocity =[v]= \red{ [ M^0 L^1 T^{-1}]}$

$\sf\longrightarrow Dimensional \ formula \ of \ Displacement =[s] =\red{ [ M^0 L^1 T^{0}]}$

$\sf\longrightarrow Dimensional \ formula \ of \ accl^n=[a]=\red{ [ M^0 L^1 T^{-2}]}$

$\underline{\purple{\textsf{Hence here we have in LHS as , }}}$

$\sf:\implies \pink{[v] = [ M^0 L^1 T^{-1}]} \\\\\sf:\implies [v]^2 = [ M^0 L^1 T^{-1}]^2 \\\\\sf:\implies \boxed{\pink{\sf [v] ^2 = [ M^0 L^2 T^{-2}]}}$

$\rule{200}2$

$\underline{\purple{\textsf{Hence here we have in RHS as , }}}$

$\sf:\implies [ u]^2 + [2as] = [ M^0 L^2 T^{-2} ] + [ M^0L^1T^{-2}][M^0L^1T^0] \\\\\sf:\implies [ u]^2 + [2as] = [ M^0 L^2 T^{-2} ] + [ M^{(0+0)}L^{(1+1)}T^{(-2+0)}] \\\\\sf:\implies \boxed{\pink{\sf [u]^2+[2as]= [ M^0 L^2 T^{-2} ] + [ M^0 L^2 T^{-2} ]}}$

$\textsf{ \to Since the dimensions of LHS and RHS is same,}\\\textsf{ i.e.\red{[ \sf M^0L^2T^{-2} ]} .Hence the formula is dimensionally correct.}$

$\underline{\underline{\textsf{\textbf{\blue{ Hence \ Proved \ ! }}}}}$