Two equal masses `m_(1) = m_(2)` = 1 kg are kept on an inclined plane with an initial separation of 0.4 m. The coefficients of friction `mu_(1)` and `mu_(2)` for the blocks `m_(1)` and `m_(2)` with ground are 0.1 and 0.2 respectively. If the collision between the two blocks is completely inelastic.

Options

A. The loss in kinetic energy is 0.16 J.
B. The loss in kinetic energy is 0.2 J.
C. The speed of `m_(1)` just after collision is 4.8 m/s.
D. The speed of `m_(2)` just after collision is 5 m/s.

Answer

A, C

Explanation

with respect to wedge
`x = h cot theta`
`a = g sin theta cos theta`
`t = sqrt((2x)/(a))=sqrt((2h)/(gsin^(2)theta))`
At time t
Velocity of wedge `=(g sin theta) t = sqrt(2gh)`
Velocity of P.w.r.t. wedge = `(g sin theta.cos theta)sqrt((2h)/(g sin^(2)theta))=(sqrt(2gh))costheta`
`V_("net")=2ghsqrt(1+cos^(2)theta+2cos(pi-theta)costheta)`
`=2ghsqrt(1+cos^(2)theta-2cos^(2)theta)`
`=2ghsqrt((1-cos^(2)theta))=2gh sin theta`