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 ⇒ (