A wedge with mass M rests on a frictionless, horizontal tabletop. A block with mass is placed on the wedge (Fig. P5.112a). There is no friction between the block and the wedge. The system is released from rest.

(a) Calculate the acceleration of the wedge and the horizontal and vertical components of the acceleration of the block.

(b) Do your answers to part (a) reduce to the correct results when M is very large?

(c) As seen by a stationary observer, what is the shape of the trajectory of the block?

Mass of the wedge is M.

Mass of the block is m.

According to Newton's second law of motion the net force on a body is equal to its mass times acceleration.

Let the acceleration of the wedge be $a_m$which only has a horizontal component. Let the acceleration of the block be $a_m$which has both horizontal and vertical components. Let N be the normal force on the block by the wedge. For the wedge the second law states

$Nsin\alpha =M{a}_M$ ..... (1)

The second law is applied on the block separately for the horizontal and vertical components. For the vertical component

$mg-Ncos\alpha =m{a}_{m}sin\alpha$ ..... (2)

For the vertical component:

$Nsin\alpha =m(-a_{m}cos\alpha +a_M)$ ..... (3)

From equations (1) and (3), you obtain

$$\begin{gathered} M{a}_M=m(-a_{m}cos\alpha +a_M) \\ {a}_M(M-m)=-m{a}_{m}cos\alpha \end{gathered}$$

$a_M=\frac{-m{a}_m\cos \alpha }{\left(M-m\right)}$ ….. (4)

From equations (1) and (2), you get

localid="1668058783182" $$\begin{gathered} \tan \alpha =\frac{M{a}_M}{m{a}_m\sin \alpha +mg} \\ {a}_M=\frac{\tan \alpha \left(m{a}_m\sin \alpha +mg\right)}{M} \end{gathered}$$

$a_M=\frac{\tan \alpha \left(m{a}_m\sin \alpha +mg\right)}{M}$ ….. (5)

Equating equations (4) and (5) and you have,

$$\begin{gathered} a_M=\frac{mg\sin \alpha \cos \alpha }{M+m{\sin }^2\alpha } \\ =\frac{\left(m+M\right)g\sin \alpha }{M+m{\sin }^2\alpha } \end{gathered}$$

The horizontal and vertical components of $a_m$are,

$$\begin{gathered} a_{m}cos\alpha -a_M=\frac{Mg\sin \alpha \cos \alpha }{M+msi{n}^2\alpha } \\ {a}_{m}sin\alpha =\frac{(m+M)gsi{n}^2\alpha }{M+msin{}^2\alpha } \end{gathered}$$

Thus, the acceleration of the wedge is localid="1668058396863" $\frac{mg\sin \alpha \cos \alpha }{M+msin{}^2\alpha }$, the horizontal component of the block acceleration is $\frac{mg\sin \alpha \cos \alpha }{M+msin{}^2\alpha }$and the vertical component is $\frac{(m+M)gsi{n}^2\alpha }{M+msin{}^2\alpha }$.

For large M, that is $M\to \infty$, the wedge acceleration becomes

$$\begin{gathered} a_{M,M\to \infty }=\frac{mg\sin \alpha \cos \alpha }{M+m{\sin }^2{\alpha }_{M\to \infty }} \\ =0 \end{gathered}$$

The horizontal component of the block acceleration becomes

$$\begin{gathered} {\left(a_m\cos \alpha -a_M\right)}_{M\to \infty }={\left(\frac{Mg\sin \alpha \cos \alpha }{M+m{\sin }^2\alpha }\right)}_{M\to \infty } \\ =g\sin \alpha \cos \alpha \end{gathered}$$

the vertical component of the block acceleration becomes

$$\begin{gathered} {\left(a_m\sin \alpha \right)}_{M\to \infty }={\left(\frac{\left(m+M\right)g{\sin }^2\alpha }{M+m{\sin }^2\alpha }\right)}_{M\to \infty } \\ =g{\sin }^2\alpha \end{gathered}$$

Thus, the total acceleration of the block is

$a_m=gsin\alpha$

For large mass of the wedge the block acceleration becomes $g\sin \alpha$and the wedge acceleration becomes 0 which are correct.

For large Mthe trajectory of the block would have been along the slope of the wedge. But when the mass of the wedge is not large, the wedge acceleration comes into effect. The acceleration of the block becomes

${\overrightarrow{a}}_m-{\overrightarrow{a}}_M$

Thus, as seen by a stationary observer, the blocks moves down with a slope larger than the slope of the wedge.