Newton's Laws of Motion Problem

[rajumahtora]

Homework Statement

Mass of Block A = 1 kg
Mass of Wedge B = 2 kg
Inclination of Wedge is 37°
and Cos 37° = 4/5
The wedge is given an acceleration of 5 ms^-2
Find -
1. Acceleration of Block wrt Wedge
2. Normal Force

Homework Equations

Newton's Law for a System (hereby referred as NLS)
∑f_ext = m₁a₁+m₂a₂...+m_na_n

The Attempt at a Solution

Let the Acceleration of Block wrt Wedge be λ along the inclination.
In Y axis, there is only one Ext force i.e. The gravity and ∑f_net in Y direction is m₁g+ m₂g = (m₁+ m₂)g
Therefore in Y axis using Newton's Law for a System,
m_Aa_A + m_Ba_B = (m₁+ m₂)g
but in Y axis , a_A = 0
and a_b in Y axis = λsinθ + a_Acos 90
= λsinθ + 0
= λsinθ
Therefore, m_Ba_B = (m₁+ m₂)g
1 (λsin(θ)) = (1 + 2)g
λ sin 37° = (3)10
λ(3/5) = 30
λ = 50
but using pseudo force, λ = 2 ms^-2 which is correct answer in the Answer Key
Where am I going Wrong?
Please Help

 

[paisiello2]

Is wedge B being given an acceleration horizontally to the left or right?

Assuming then no friction block A will also be sliding down the inclined wedge B.

 

[haruspex]

in Y axis using Newton's Law for a System,
m_Aa_A + m_Ba_B = (m₁+ m₂)g

Are you assuming that the blocks are in free fall? I would have guessed the wedge is supported somehow.