Question

Block A of mass m is placed over a wedge B of same mass m. Assuming all surfaces to be smooth. The displacement of block A in 1 s if the system is released from rest is

​

Answer 1

Given: Block A of mass m is placed over a wedge B of same mass m.

To find: The displacement of block A in 1 s.

Solution:

• Now we have given block A of mass m is placed over a wedge B of same mass m.

• Now acceleration of A w.r.t ground will be:

             a(A/g) = a(A/B) + a(B/g)             (A/g = accel. of A w.r.t ground)

• So in x direction, it will be:

             a(A/g)x = a(A/B)x + a(B/g)x        ............(i)

• From fbd, a(A/g)x = 0 which implies a(A/B)x = -a(B/g)x

• So for block A, we have:

             mg - N = m ( a sin theta)  ...............(ii)

• For block B, we have:

             (N +mg) sin theta = ma .................(iii)

• Now acceleration of A is:

             a(A) = a sin theta

                    = (2g sin theta / 1 + g sin^2 theta ) sin theta   ....from (ii) and (iii)

• Now displacement in 1 sec, we have:

             S = ut + 1/2at^2

             S = 0 + 1/2 (  (2g sin theta / 1 + sin^2 theta ) sin theta)(1)^2

             S = (g sin^2 theta / 1 + sin^2 theta )

Answer:

       So the displacement is (g sin^2 theta / 1 + sin^2 theta )