A block of mass M is placed on the top of a bigger triangular block of mass 10M with base of 2.2m?

Question

All the surfaces are frictionless .the system is released from rest.Prove that the distance moved by the bigger block at the instant when the smaller block reaches the ground is 0.2m.
(shape of block is like that of a right angled triangle)

Answer 1

The normal force N applied to the sliding surface by a block of mass m is

N=mg cos(A)

A- angle the hypotenuse of the larger mass makes with the horizontal
m- mass of the block
g - acceleration due to gravity

The force that the small block pushes the large block along the horizontal (and propels itself as well)
F= N cos(B) B= 90-A or pi/2 - A
F=mgcos(A) cos(B)

The distance lower block will slide
S= 0.5 at^2
a=F/(m+10m) then
S=F t^2 / 2(m+10m)]

It takes the same time for the 10m to slide horizontally as it takes the smaller block to slide along the side

t^2= 2C/ (gsin(A))
gsin(A) is the free fall acceleration component along the inclined.
C is the length of the inclined
Substituting we have
S=F 2C/ (2gsin(A)) (m+10m)]
S=F C/ (gsin(A)) (m+10m)]
also F=mgcos(A) cos(B)
S=mgcos(A) cos(B) C/[(gsin(A)) (11m)]
S=cos(A) cos(B) C/[(sin(A)) (11)]
Now C= base/ cos(A)
S=cos(A) cos(B) base /[(cos(A)sin(A)) (11)]
so
S= (base/11) (cos(B) / sin(A) )

if the triangle is also equaleteral A=B
S=base/11=2.3/11=0.2m