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In this article, we will find the dimension of Impulse<\/p>\n\n\n\n
Dimensional formula for Impulse is<\/p>\n\n\n\n
![\"Dimension](\"https:\/\/physicscatalyst.com\/article\/wp-content\/uploads\/2019\/12\/dimension-of-impulse.png\")
<\/figure><\/div>\n\n\n\nWhere L<\/strong> -> Length<\/p>\n\n\n\n T-> Time Impulse is defined as the product of Force and time. It is denoted by letter I Now dimension of Time is given by $[T]$ Dimension of Mass is given by $[M]$ So Dimension of Acceleration is given by Now from equation (1) , we can determine the dimension of Impulse as $= [M^1L^1T^{-2}] \\times [T^1] = [M^1L^1T^{-1}] $ <\/p>\n\n\n\n Unit of Impulse is Kg m\/s <\/p>\n\n\n\n Try the free Quiz given below to check your knowledge of Dimension Analysis:- <\/p>\n\n\n\n
M<\/strong> -> Mass<\/p>\n\n\n\n
We would now derive this dimensional formula. <\/p>\n\n\n\n Derivation for expression of Dimension of Impulse<\/h2>\n\n\n\n
$I= F \\times t$ -(1)
Where I = Impulse
F= Force
t=Time<\/p>\n\n\n\n
Lets derive the dimension of Force
Force is defined as product of Mass and acceleration
$F= m \\times a$<\/p>\n\n\n\n
Dimension of Acceleration is derived as
$a= \\frac {dv}{dt}$
Where v = velocity, t= time
Lets derive the dimension of Velocity
$v= \\frac {dx}{dt}$
or
$v = \\frac {d}{t}$
Now
Where
d-> displacement
t -> Time
Now Dimension of Displacement = $[L^1]$
Therefore, dimension of Velocity= $\\frac {[L^1]}{[T^1]}=[M^0 L^1T^{-1}]$ <\/p>\n\n\n\n
$\\text{Dimension of Acceleration} =\\frac {\\text{dimension of velocity}}{\\text {dimension of time}}$
$\\text{Dimension of Acceleration} =\\frac { [M^0 L^1T^{-1}] }{[T]}= [M^0L^1T^{-2}]$
Hence Dimension of Force will be
$ \\text{Dimension of Force} = [M^1 ] \\times [ M^0L^1T^{-2}] = [M^1L^1T^{-2}]$ <\/p>\n\n\n\n
$ \\text{Dimension of Impulse} = (\\text{Dimension of force}) \\times (\\text{Dimension of time}) $ <\/p>\n\n\n\n
\nQuiz on Dimensional Analysis<\/strong><\/span><\/h2>\n
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