CGPET 2016 Solved Paper

Given,
Mas of block = 10kg
Mass of bullet = 50kg
=50×10−3‌kg
=5×10−2‌kg
m=50
v=?

Speed of bullet =V
Distance covered by the block =s=2m
Coefficient of friction between the block and the table μ=0.05
Thus,
F=Ff,N=Mg
⇒Ma=μN= acceleration of block
⇒ Ma=μ‌Mg=μN
⇒ a=0.05×10=0.05∕s2
Now, for speed of block
vblock2=u2+2‌as
=0+2‌as=2×0.5×2=2m∕s2
⇒vblock=√2
From law of conservation of momentum
mv=Mvblock
⇒ v=

10×√2

50×10−3

=

√2

5

×103=2√2×102
Now for freely falling body
Final velocity =

v

10

Initial velocity u=0
Using formula v2=u2=2‌gH
⇒ (

v

10

)2=−2‌gH
⇒ [

2√2×102

10

]2=−2‌gH
⇒ −2‌gH=(20√2)2=400×2=800
⇒H=

800

2

×10(=)40M=40×10−3‌km
=0.04‌km