Laws of MotionHard
Question
A block B of mass 0.6 kg slides down the smooth face PR of a wedge A of mass 1.7 kg which can move freely on a smooth horizontal surface. The inclination of the face PR to the horizontal is 45o. Then :
Options
A.the acceleration of A is 3 g/20
B.the vertical component of the acceleration of B is 23 g/40
C.the horizontal component of the acceleration of B is 17 g/40
D.none of these
Solution
F.B.D. of block B w.r.t. wedge
for block A
N cos 45o = 1.7 a .....(i)
for block B
0.6g sin 45o + 0.6a cos 45o = 0.6b .....(ii)
N + 0.6 a cos 45o = 0.6 g cos 45o .....(iii)
by solving (i), (ii) & (iii)
a =
and b = 
Now vertical component of acceleration of B = b cos 45o =
and horizontal component of acceleration of B = b sin 45o − a =
for block A
N cos 45o = 1.7 a .....(i)
for block B
0.6g sin 45o + 0.6a cos 45o = 0.6b .....(ii)
N + 0.6 a cos 45o = 0.6 g cos 45o .....(iii)
by solving (i), (ii) & (iii)
a =
and b = Now vertical component of acceleration of B = b cos 45o =
and horizontal component of acceleration of B = b sin 45o − a =
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