Question

Figure shows a wedge on which a small block is released from rest. All the surfaces are system comprises of wedge and blocks. Mark the correct statement(s) regarding motion of wedge till block attains maximum height on wedge.

• A. Acceleration of centre of mass of system is initially vertically down then vertically up.
• B. Initially centre of mass moves down and then up.
• C. At the maximum height block and wedge move with common velocity.
• D. Centre of mass of wedge moves towards left then right

Answer 1

Answer: (B), (C)

B, C

Answer 2

The problem describes a small block released from rest on a smooth wedge, forming a system. We need to analyze the motion of the wedge until the block attains its maximum height on the wedge. Since all surfaces are smooth, there is no friction.

System Definition and Forces: The system consists of the block (mass $m$) and the wedge (mass $M$). External forces acting on the system are:

1. Gravity on the block ($mg$), vertically downwards.
2. Gravity on the wedge ($Mg$), vertically downwards.
3. Normal force from the ground on the wedge ($N_{ground}$), vertically upwards.

Horizontal Motion Analysis: There are no external horizontal forces acting on the system. Therefore, the total horizontal momentum of the system (block + wedge) is conserved. Initially, the system is at rest, so the initial horizontal momentum is zero. $P_{x, initial} = m(0) + M(0) = 0$. Since horizontal momentum is conserved, the horizontal velocity of the center of mass of the system ($V_{CM,x}$) must remain zero. Consequently, the horizontal acceleration of the center of mass of the system ($a_{CM,x}$) is also zero.

Vertical Motion Analysis: The external vertical forces are $mg$, $Mg$, and $N_{ground}$. The vertical acceleration of the center of mass is given by $a_{CM,y} = \frac{\sum F_{ext,y}}{m+M} = \frac{N_{ground} - (m+M)g}{m+M}$. Alternatively, $a_{CM,y} = \frac{m a_{by} + M a_{wy}}{m+M}$. Since the wedge moves horizontally, $a_{wy}=0$. So, $a_{CM,y} = \frac{m a_{by}}{m+M}$, where $a_{by}$ is the vertical acceleration of the block.

(A) Acceleration of centre of mass of system is initially vertically down then vertically up. When the block is released from rest on the right side of the wedge, it starts sliding downwards. So, its initial vertical acceleration ($a_{by}$) is downwards (negative). Therefore, the initial acceleration of the center of mass of the system ($a_{CM,y} = \frac{m a_{by}}{m+M}$) is vertically downwards. As the block moves up the left side of the wedge, its vertical velocity is upwards, but it is decelerating (losing vertical speed). This means its vertical acceleration ($a_{by}$) is still downwards (negative). Thus, throughout the motion until the block reaches its maximum height on the wedge (where $v_{by}=0$), the vertical acceleration of the block $a_{by}$ is generally downwards (or zero if it were at a horizontal tangent point). Consequently, $a_{CM,y}$ is always vertically downwards. Therefore, statement (A) is incorrect.

(B) Initially centre of mass moves down and then up. The vertical velocity of the center of mass of the system is $V_{CM,y} = \frac{m v_{by} + M v_{wy}}{m+M}$. Since $v_{wy}=0$ (wedge moves horizontally), $V_{CM,y} = \frac{m v_{by}}{m+M}$. The block starts from rest at some height. As it slides down the wedge, its vertical velocity ($v_{by}$) is negative (downwards). So, $V_{CM,y}$ is initially negative (downwards). After reaching the lowest point, the block moves up the other side of the wedge. During this phase, its vertical velocity ($v_{by}$) is positive (upwards). So, $V_{CM,y}$ becomes positive (upwards). At the maximum height on the wedge, the block's vertical velocity ($v_{by}$) becomes zero, so $V_{CM,y}$ becomes zero. Thus, the center of mass of the system initially moves down, then moves up, and momentarily stops vertically at the maximum height. Therefore, statement (B) is correct.

(C) At the maximum height block and wedge move with common velocity. "Attaining maximum height on wedge" implies that the block momentarily stops its vertical motion relative to the wedge, i.e., its vertical velocity relative to the ground ($v_{by}$) is zero. Furthermore, for the block to be at a "maximum height" (a turning point in its motion on the wedge), its velocity relative to the wedge must momentarily be zero. If there were still relative horizontal motion, the block would continue to slide on the wedge. Thus, at this point, the block's velocity relative to the wedge is zero. This means the block's velocity relative to the ground ($v_b$) is equal to the wedge's velocity relative to the ground ($V_w$). So, $v_b = V_w$. Since horizontal momentum is conserved for the system: $m v_b + M V_w = 0$ Substituting $v_b = V_w$: $m V_w + M V_w = 0$ $(m+M) V_w = 0$ Since $(m+M) \ne 0$, it implies $V_w = 0$. If $V_w=0$, then $v_b=0$. This means that at the maximum height, both the block and the wedge are momentarily at rest. This is consistent with the conservation of mechanical energy, as the block would return to its initial height, converting all its potential energy back to potential energy. Therefore, statement (C) is correct.

(D) Centre of mass of wedge moves towards left then right. Let's consider the horizontal forces and momentum. When the block is released from the right side, it slides down and moves towards the left (relative to the ground). To conserve horizontal momentum ($m v_{bx} + M V_w = 0$), if $v_{bx}$ is negative (leftward), then $V_w$ must be positive (rightward). This means the wedge moves to the right. This can also be seen by considering the horizontal component of the normal force. As the block slides down the right side, the normal force exerted by the block on the wedge has a horizontal component pointing to the right, causing the wedge to accelerate to the right. Then, as the block moves up the left side, it continues to move to the left (relative to the ground) until it momentarily stops at the maximum height. During this phase, the normal force exerted by the block on the wedge has a horizontal component pointing to the right, continuing to accelerate the wedge to the right. Therefore, the wedge moves only towards the right, not left then right. (Note: If the problem implies the block starts on the left and moves right, then the wedge would move left. But based on the figure, it starts on the right and moves left.) Therefore, statement (D) is incorrect.