Center of MassHard
Question
A bullet of mass m moving vertically upwards instantaneously with a velocity ′u′ hits the hanging block of mass ′m′ and gets embedded in it. As shown in the figure the height through which block rises after the collision, assume sufficient space above block :
Options
A.u2/2g
B.u2/g
C.u2/8g
D.u2/4g
Solution
From momentum conservation
mu = 2mv
⇒ v =
from energy conservation
× 2m × 
= 2 mgh
⇒ h =
mu = 2mv
⇒ v =
from energy conservation
× 2m × = 2 mgh⇒ h =
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