In a direct impact, loss in kinetic energy is given by
$Δ K = \frac{M_{1} M_{2}}{2 (M_{1} + M_{2})} . (V_{1} - V_{2})^{2} (1 - k^{2})$
with usual notations (except k). The quantity k will have dimensional formula:

$[M^{0} L^{2} T^{- 2}]$

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$[M L T^{- 1}]$

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$[M^{0} L^{0} T^{0}]$

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$[M^{0} L^{1} T^{- 1}]$

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Solution

The correct option is C
$[M^{0} L^{0} T^{0}]$

Dimension of $k^{2}$ will be same as that of 1, i.e., $[M]^{0} [L]^{0} [T]^{0}$ .
Hence, dimension of $k$ is $[M]^{0} [L]^{0} [T]^{0}$ .

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Similar questions

In a direct impact, loss in kinetic energy is given by $△ K = \frac{M_{1} M_{2}}{2 (M_{1} + M_{2})} . (V_{1} - V_{2})^{2} (1 - K^{2})$ with usual

notations(except k). The quality k will have dimensional formula:

Δ K = \frac{m_{1} m_{2}}{2 (m_{1} + m_{2})} | {\to u}_{1} - {\to u}_{2} |^{2} (1 - k^{2})

k

k

k = C x

C

x

C

[M L T^{- 1}]

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