A block of mass m is placed at rest on the top of a smooth wedge of mass M, which in turn is placed at rest on a smooth horizontal surface as shown in figure. Then the distance moved by the wedge as the block reaches the foot of the wedge is
A
(M+mm)l
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(Mm+M)l
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(mM+m)l
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
(M+mM)l
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C(mM+m)l Initially both wedge and block are at rest.
There is no external force in horizontal direction of the wedge block system.
Thus acceleration, ⇒acom in x direction is zero.
Therefore, x− coordinate of the COM of the system will remain at rest even when block slides down.
Assume that wedge moves X when block reaches the ground.
As xcom is not changing. m1x1=m2x2