Question

# A bullet of mass m moving horizontally with a velocity v strikes a block of wood of mass M and gets embedded in the block. The block is suspended from the ceiling by a massless string. The height to which block rises is:

A
v22g(mM+m)2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
v22g(M+mm)2
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
v22g(mM)2
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
v22g(Mm)2
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is A v22g(mM+m)2The situation is as shown in the figure.Let V be velocity of the block-bullet system just after collision. Then by the law of conservation of linear momentum, we getmv=(m+M)VV=mvm+MLet the block rises to a height h.According to law of conservation of mechanical energy,we get12(m+M)V2=(m+M)ghh=V22g=v22g(mm+M)2.

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program