In head on collision of two point particles, loss in kinetic energy is given by Δ K=(m1 m2/2(m1+m2))| vecu1- vecu2|2(1-k2) With usual notations (except k ), the dimensional formula of quantity k is

Question

In head on collision of two point particles, loss in kinetic energy is given by Δ K=(m1 m2/2(m1+m2))| vecu1- vecu2|2(1-k2) With usual notations (except k ), the dimensional formula of quantity k is (A) [ M 0 L 0 T -1] (B) [ M 0 L 0 T 0] (C) [ M 0 LT -1] (D) [ M 0 L 2 T -2]

Tardigrade Admin · Accepted Answer

It is clear that (1- k 2) is dimensionless So [ k ]=[ M 0 L 0 T 0]

Q. In head on collision of two point particles, loss in kinetic energy is given by
$Δ K = \frac{m _{1} m _{2}}{2 ( m _{1} + m _{2} )} ∣ u_{1} - u_{2} ∣^{2} (1 - k^{2})$
With usual notations (except $k$ ), the dimensional formula of quantity $k$ is

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

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

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

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

It is clear that $(1 - k^{2})$ is dimensionless
So $[k] = [M^{0} L^{0} T^{0}]$

Q. In head on collision of two point particles, loss in kinetic energy is given by ΔK=2(m1​+m2​)m1​m2​​∣u1​−u2​∣2(1−k2) With usual notations (except k ), the dimensional formula of quantity k is

[M0L0T−1]

[M0L0T0]

[M0LT−1]

[M0L2T−2]