Question

Kinetic energy of a particle (K.E) moving along x-axis depends on its position (x) and time (t) as KE = At^2/B+x^1/2 +Cx^3/2 / D+t^2 . The dimensional formula of ABC/D is

Answer 1

The displacement of a particle at time t is given s = ut + 1/2at2

At time (t - 1), the displacement of a particle is given by

S' = u (t-1) + 1/2a(t-1)2

So, Displacement in the last 1 second is,

St = S - S'

= ut + 1/2 at2 – [u(t-1)+1/2 a(t-1)2 ]

= ut + 1/2at2 - ut + u - 1/2a(t - 1)2

= 1/2at2 + u - 1/2 a (t+1-2t) =  1/2at2 + u - 1/2at2 - a/2 + at

S = u + a/2(2t - 1)

(b) Putting the values of u = 2 m/s, a = 1 m/s2 and t = 5 sec, we get

S = 2 + 1/2(2 x 5 - 1) = 2 + 1/2 x 9

= 2 + 4.5 = 6.5 m  

Answer 2

The correct answer is  $M^{2} L^{\frac{7}{2} } T^{-6}$.

Given: KE = $\frac{At^{2} }{B+x^{\frac{3}{2} } } +\frac{Cx^{\frac{3}{2} } }{D+t^{2} }$

To Find: Dimensions of  $\frac{ABC}{D}$.

Solution:

Two items can only be added or subtracted if their dimensions are same.

So dimension of B = $x^{\frac{3}{2} } =L^{\frac{3}{2} }$.

Dimension of KE = $M^{1} L^{2}T^{-2}$.

Dimension of  $\frac{At^{2} }{B+x^{\frac{3}{2} } }$ = $M^{1} L^{2}T^{-2}$.

$\frac{A[T]^{2} }{L^{\frac{3}{2} } }$  = $M^{1} L^{2}T^{-2}$.

A = $M^{1} L^{\frac{7}{2} }T^{-4}$

Dimension of D = $t^{2} =T^{2}$.

Dimension of $\frac{Cx\frac{3}{2} }{D+t^{2} } =M^{1}L^{2} T^{-2}$

$\frac{C[L]^{\frac{3}{2} } }{T^{2} } =M^{1}L^{2} T^{-2}$

Dimension of C = $M^{1}L^{\frac{1}{2} }$

So  

$\frac{ABC}{D}$ = $M^{2} L^{\frac{7}{2} } T^{-6}$

Hence, the value of  $\frac{ABC}{D}$ is  $M^{2} L^{\frac{7}{2} } T^{-6}$.

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