Pulley system on a frictionless cart

Yes, the cart will move, due to the force applied by the string to the pulley.

To solve, calculate the string tension while the weights are moving, and then note that the pulley has to provide an opposing force in order to change the string's direction. The reaction to that force acts upon the cart, accelerating it.

Momentum is conserved because the resting weight is accelerated to the right, while the cart is accelerated to the left.

Calculating the actual numbers will be entertaining, as you must include the cart's acceleration while calculating the string tension. I'm guessing that including a new, accelerating reference frame won't be helpful, as you won't know the magnitude of the acceleration until the problem has been solved.

Edit: as noted in a comment by dmmckee, the answer will depend on whether the hanging weight is constrained to stay in contact with the cart, or is free to swing away from it (which it would do if allowed).

Let the mass of cart be $M$, mass of hanging weight be $m_1$ and mass on top of cart be $m_2$. Choosing and inertial coordinate system, let the x coordinate of $M$ , $m_1$ and $m_2$ be $X$ , $x_1$ and $x_2$ respectively. Let $T$ be the tension in the string and $a$ be the acceleration of the masses ( horizontal for $m_2$ and vertical for $m_1$).

Equations of motion give:

$$ m_1g -T=m_1a $$ $$ T=m_2a \tag{0}$$ Hence, $a=m_1g/(m_1+m_2) \tag{1}$

let $X_{cm}$ be the X centre of mass coordinate of the system. $$X_{cm}=\frac{m_1x_1+m_2x_2+MX}{m_1+m_2+M}$$

differentiating twice; $$\ddot{X}_{cm}=\frac{m_1\ddot{x}_1+m_2\ddot{x}_2+M\ddot{X}}{m_1+m_2+M} \tag{2}$$ Since $\ddot{X}_{cm} =0 $ , $\ddot{x}_1=\ddot{X}$,($m_1$ does not swing ) and $\ddot{x}_2=a$ , $$m_1\ddot{X} + m_2a+ M\ddot{X} = 0 \tag{3}$$ We can use eq(1) and eq(3) to find $X, x_1, x_2$ as a function of time.

Method 2:

Since @Buraian wants equations with the method @Daniel Griscom suggested, here they are: Consider the part of the string that is in contact with the pulley. It experiences a force $T$ downwards and $T$ towards the left. Say the pulley applies a force of $N_1$ on the string ( towards upper right). By Newton's third law, the string applies $N_1$ on mass $M$ (towards bottom left).

Since $m_1$ doesn't swing, the rail (part of mass $M$) applies a force $N_2$ (towards left) on $m_1$ and $m_1$ applies a force of $N_2$ on $M$ towards right. Let $M$ (and $m_1$)accelerate with $a^{'}$, horizontally.

Since the net force on a mass less string is always $0$, $$N_1cos(45^0)=T \tag{4}$$ from eq(1),eq(0) and eq(4) we can get the value of $N_1$.

Equation of motion for $M$: $$N_1cos(45^0)-N_2 = Ma^{'} \tag{5}$$ Equation of motion for $m_1$ (X direction ): $$N_2=m_1a^{'}\tag{6}$$

Needless to say, by eq(5) and eq(6) we can obtain all the quantities and predict the motions of blocks. we get $a^{'}$ which is the same as $\ddot{X}=m_1m_2g/((m_1+m_2)(m_1+M))$ from first method.