The inclined surfaces of two moveable wedges of the same mass M = 2kg are smoothly placed just in contact with each other and placed on the horizontal plane as shown in the figure. A small block of mass m = 1kg slides down the left wedge from a height h= 9m. Neglect the friction and both wedges can move independently .
(i) Find velocity of left wedge just after when block leave the wedge
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)
(ii)
(iii)
By momentum conservation
Mu = mV (i)
V = 
u
By energy conservation
mgh = 
Mu 2 + 
mV 2 = 
Mu 2 + 
m 
u 2
mgh = 
Mu 2 + 
u 2 = 
u 2 
2m 2 gh = u 2 (Mm + M 2 )
∴ 
= u 2 .
u = m 
...(ii)
By momentum conservation
mV = (M + m) V 1
V 1 = 
...(iii)
By energy conservation
mV 2 = 
(m + M) V 1 2 + mgh 1
mV 2 = 
(m + M) 
+ mgh 1
⇒ 
mV 2 – 
= mgh 1
V 2 
= mgh 1
⇒ 
= mgh 1 ...(iii)
Put V = – 
× u and u = m 
∴ V = M 
....(iv)
put value of V from eqn (iv) to (iii)
∴ h ' = 
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