A block of mass 1 kg is moving towards a movable wedge of mass 2 kg as shown. All surfaces are smooth. When block leaves the wedge from top, its velocity is making an angle θ = 30º with horizontal.
(i) The value of V 0 in m/s is–
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) : By conservation of momentum
1 × V b cos 30º + 2V w = 1 + V 0 ……(i)
Now, Net velocity of block, 
= 
+ 
also 
makes 30º with horizontal and 
is at 60º with horizontal
| 
| = | 
|
| 
| = 
= 
V w
from equation (i)
V w cos30º + 2V W = V 0
V 0 = 
V w …………. (ii)
from conservation of energy
1×10×1.45+ 
×1× 
+ 
×2× 
= 
×1 
Solving V 0 = 7 m/s
(ii) : h max = 1.45 + 
= 1.6 m
(iii) : Applying impulse momentum equation in horizontal direction. J H = |change in momentum in horizontal direction|
J H = 2V w = 2 × 
V 0 = 4 N–S
J V = |impulse of vertical component of Normal reaction|
= J H cot 60º = 
N – S
J = 
= 
N – S
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