Let us define a coordinate system with unit vectors $\hat{i}$ (horizontal, right) and $\hat{j}$ (vertical, up).

• $\vec{a}_2$: Acceleration of the wedge $M$. It moves horizontally to the right.
  $\vec{a}_2 = a_2 \hat{i}$
• $\vec{a}_1$: Absolute acceleration of the block $m$. Based on the diagram and equations, it acts at an angle $\phi$ below the horizontal (towards the left).
  $\vec{a}_1 = -a_1 \cos\phi \hat{i} – a_1 \sin\phi \hat{j}$
• $\vec{N}$: Normal force exerted by the wedge on the block. The incline angle is $\theta$. The normal acts at angle $\theta$ to the vertical (pointing up and left).

Considering the forces on the wedge in the horizontal direction ($\hat{i}$):

The only external horizontal force acting on the wedge is the horizontal component of the normal force $N$ from the block. The normal force pushes the wedge to the right.

Horizontal Component of Force: $F_x = N \sin\theta$

Using Newton’s Second Law ($F = ma$):

Considering the forces on the block in the horizontal direction:

The normal force $N$ acts on the block with a horizontal component $N \sin\theta$ directed in the $-\hat{i}$ direction (left). This force provides the horizontal acceleration.

Horizontal acceleration of block: $-a_1 \cos\phi$

The block must remain in contact with the wedge. This implies that the relative velocity (and acceleration) perpendicular to the inclined surface must be zero. In other words, the component of the block’s absolute acceleration perpendicular to the incline must equal the component of the wedge’s acceleration perpendicular to the incline.

• Unit vector perpendicular to incline (pointing up-left): $\hat{n} = -\sin\theta \hat{i} + \cos\theta \hat{j}$
• Component of $\vec{a}_2$: $a_{2,\perp} = \vec{a}_2 \cdot \hat{n} = (a_2 \hat{i}) \cdot (-\sin\theta \hat{i} + \cos\theta \hat{j}) = -a_2 \sin\theta$
• Component of $\vec{a}_1$: $a_{1,\perp} = \vec{a}_1 \cdot \hat{n} = (-a_1 \cos\phi \hat{i} – a_1 \sin\phi \hat{j}) \cdot (-\sin\theta \hat{i} + \cos\theta \hat{j})$
  $= a_1 \cos\phi \sin\theta – a_1 \sin\phi \cos\theta = -a_1 (\sin\phi \cos\theta – \cos\phi \sin\theta) = -a_1 \sin(\phi – \theta)$

Equating the magnitudes (since both point into the wedge surface):

From equations (1) and (2), both expressions equal $N \sin\theta$. Therefore:

From the constraint equation (3), we can substitute the ratio $\frac{a_2}{a_1}$:

Substituting this into the mass ratio expression:

Based on the values derived in the handwritten note:

• $\sin(\phi – \theta) = 1/2 \Rightarrow (\phi – \theta) = 30^\circ$
• $\sin\theta = 1/2 \Rightarrow \theta = 30^\circ$
• $\cos\phi = 1/2 \Rightarrow \phi = 60^\circ$ (Consistent since $60^\circ – 30^\circ = 30^\circ$)

Substituting these values: