Question

A fixed smooth inclined surface has an inclination θ with the horizontal. A wedge B of mass m is placed on the incline as shown in figure. The block A of mass m rests on the smooth horizontal surface of wedge B. If the system is released from rest, the displacement of block A in ground frame in1 second is



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Answer 1

The displacement of block A in ground frame in 1 second is  (g sin² θ)/(1 + sin² θ)

Explanation:

The image given in the question is attached below.

$\overrightarrow{a_{A/g}}=\overrightarrow{a_{A/B}}+\overrightarrow{a_{B/g}}$

$\left [ a_{A/g} \right ]_x = \left [ a_{A/B} \right ]_x+\left [ a_{B/g} \right ]_x$ → (Equation 1)

$\left [ a_{A/g} \right ]_x = 0$

⇒ $\left [ a_{A/B} \right ]_x = -\left [ a_{B/g} \right ]_x$

Block A:

mg - N = m(a sin θ) → (Equation 2)

Block B:

(N + mg) sin θ = ma → (Equation 3)

aA = a sin θ

aA = [(2g sin θ)/(1 + (g sin² θ))] sin θ

∴ aA = (2g sin² θ)/(1 + sin² θ)

Now,

S = u + 1/2at²

⇒ S = 0 + (1/2 × aA × t²)

Where,

t = 1 second

Now,

S = 1/2 × (2g sin² θ)/(1 + sin² θ) × (1)

∴ S = (g sin² θ)/(1 + sin² θ)