A wedge (free to move) of mass ‘M’ has one face making an angle α with horizontal and is resting on a smooth rigid floor. A particle of mass ‘m’ hits the inclined face of the wedge with a horizontal velocity v 0 . It is observed that the particle rebounds in vertical direction after impact. Neglect friction between particle and the wedge & take M = 2m, v 0 = 10m/s, tan α = 2, g = 10m/s 2 
Assume that the inclined face of the wedge is sufficiently long so that the particle hits the same face once more during its downward motion. Calculate the time elapsed between the two impacts.
Text Solution
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(3)
Applying momentum conservation in horizontal direction
mV 0 = Mu M = 2m
u = 
Eqn of e along normal
e = 
= 
e = 
cot α + 
...(i)
Along incline surface of wedge friction is negligible so change in momentum
mV 0 cos α = mV sin α
= cot α ...(ii)
Put value of (ii) in (i)
e = cot 2 α + 
given tan α = 2
= 
h = (ut) tan α
By (2)nd eq. of motion
– h = Vt – 
gt 2
– (ut) tan α = Vt – 
gt 2
Or – u tan α = V – 
gt
gt = V + u tan α
t = 
(V 0 cot α + 
tan α )
t = 
V 0 tan α (cot 2 α + 
)
t = 
(e) = 
substituting values : 
= 3sec
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