The correct option is
A v=0 When both blocks reach the ground, let their speeds be
v1 each and let the velocity of wedge be
v towards right.
For mass
m1=m, which is sliding towards left,
v1x=−v1cos60∘ For mass
m2=m which is sliding towards right,
v2x=v1cos60∘ For wedge
m3=10m which is sliding in
x direction,
v3x=v There is no external force in the
x-direction, so velocity of
(COM)x will be zero. (because, initially the system was at rest)
i.e
Vcom=m1v1x+m2v2x+m3v3xm1+m2+m3=0 ⇒m(−v1cos60o)+m(v1cos60o)+10m(v)m+m+10m=0 ⇒v=0 Hence, velocity of wedge will be zero.
Since there is no external force in
x direction, we can say
(acom)x will be zero. Hence, acceleration for all blocks is
g along the
y direction
acom=m1a1+m2a2+m3a3m1+m2+m3 ⇒acom=m(−g)+m(−g)+10m(−g)m+m+10m ⇒acom=g (downward)
i.e
acom=−g^j