A block A of mass 2m is placed on another block B of mass 4m which in turn is placed on a fixed tabl

Q: Why doesn't block A move when block B is struck?

Because there is absolutely no friction between block A and block B. Without friction, there is no horizontal force to transfer momentum from B to A. Block A simply stays in its place due to inertia.

Q: Why is the toppling condition $s > 2d$?

Block A has a total length of $4d$. Assuming uniform mass distribution, its center of mass is exactly in the middle, at a distance of $2d$ from its left edge. For block A to topple, the supporting block B must slide completely past this center of mass.

Q: Why do we use $e=1$ for the collision?

The problem states that the collision is "elastic". In a perfectly elastic collision, the relative velocity of separation equals the relative velocity of approach, which mathematically means the coefficient of restitution $e$ is 1.

Q: Why is the normal force $6mg$ and not $4mg$?

Even though block A doesn't move horizontally, it is still resting on block B. The table must support the weight of both blocks. Therefore, the total normal force exerted by the table is the sum of their weights: $(4m + 2m)g = 6mg$.

Q: Why does the small mass rebound?

When a lighter object collides elastically with a much heavier object (in this case, $m$ hitting $4m$), the lighter object tends to bounce back. The math confirms this: solving the momentum and restitution equations gives a negative final velocity for the small mass.

Question

A block $A$ of mass $2m$ is placed on another block $B$ of mass $4m$ which in turn is placed on a fixed table. The two blocks have a same length $4d$ and they are placed as shown in figure. The coefficient of friction (both static and kinetic) between the block $B$ and table is $\mu$. There is no friction between the two blocks. A small object of mass $m$ moving horizontally along a line passing through the centre of mass (CM) of the block $B$ and perpendicular to its face with a speed $v$ collides elastically with the block $B$ at a height $d$ above the table.
(a) What is the minimum value of $v$ (call it $v_0$) required to make the block $A$ to topple?
(b) If $v = 2 v_0$, find the distance (from the point $P$ in the figure) at which the mass $m$ falls on the table after collision. (Ignore the role of friction during the collision.)

Sigma Percentile · Accepted Answer

The problem presents a fascinating interplay of collisions, friction, and rigid body dynamics. We have a two-block system resting on a table, with a small mass acting as the trigger. Let's break down the physics step by step.

### The Setup: A Frictionless Trap

Imagine block $B$ of mass $4m$ resting on a table with friction coefficient $\mu$. On top of it sits block $A$ of mass $2m$. The crucial detail here is the **absence of friction** between $A$ and $B$. This means block $A$ is completely isolated from any horizontal forces. When a small mass $m$ strikes block $B$ with velocity $v$, block $B$ will be jolted into motion, but block $A$ will simply try to stay exactly where it is due to inertia.

### The Elastic Collision: Exchanging Momentum

The small mass $m$ collides elastically with block $B$. In an elastic collision, both linear momentum and kinetic energy are conserved. We can express this using the momentum equation and the coefficient of restitution ($e=1$).

Let $v_1$ be the final velocity of the small mass and $v_2$ be the final velocity of block $B$. 
Conservation of momentum gives us:
$$mv = mv_1 + 4mv_2 \implies v = v_1 + 4v_2$$

The coefficient of restitution equation (velocity of separation equals velocity of approach) gives:
$$v = v_2 - v_1$$

Solving these two linear equations simultaneously, we add them to eliminate $v_1$:
$$2v = 5v_2 \implies v_2 = \frac{2}{5}v$$

Substituting $v_2$ back to find $v_1$:
$$v_1 = \frac{2}{5}v - v = -\frac{3}{5}v$$

The negative sign indicates that the small mass rebounds to the left, while block $B$ moves to the right.

### The Deceleration: Friction Takes Over

As block $B$ slides to the right, it experiences kinetic friction from the table. To find this frictional force, we must consider the total normal reaction. Even though block $A$ isn't moving horizontally, its weight still presses down on block $B$.

The total normal force is $N = (4m + 2m)g = 6mg$.
The frictional force is $f_k = \mu N = 6\mu mg$.

This friction causes block $B$ to decelerate. Using Newton's second law, the deceleration $a$ is:
$$a = \frac{f_k}{4m} = \frac{6\mu mg}{4m} = \frac{3}{2}\mu g$$

Using the third equation of motion ($v_f^2 = v_i^2 - 2as$), we can find the stopping distance $s$ of block $B$:
$$0 = \left(\frac{2}{5}v\right)^2 - 2\left(\frac{3}{2}\mu g\right)s$$
$$s = \frac{4v^2/25}{3\mu g} = \frac{4v^2}{75\mu g}$$

### The Toppling Condition: Losing Support

Now, how does block $A$ topple? Since block $A$ has length $4d$ and is aligned with the left edge of block $B$, its center of mass (COM) is located at a distance of $2d$ from the left edge.

Because there is no friction, block $A$ remains stationary relative to the ground. Block $B$ is sliding out from underneath it. For block $A$ to topple, the left edge of block $B$ must slide completely past the COM of block $A$. Therefore, block $B$ must move a distance greater than $2d$.

$$s > 2d$$
$$\frac{4v^2}{75\mu g} > 2d \implies v^2 > \frac{150\mu gd}{4} = \frac{75}{2}\mu gd$$

Taking the square root, we find the minimum initial velocity $v_0$:
$$v_0 = \frac{5}{2}\sqrt{6\mu gd}$$

### The Rebound: Projectile Motion

For part (b), we are given an initial velocity $v = 2v_0 = 5\sqrt{6\mu gd}$. 
We already know the small mass rebounds with a speed of $|v_1| = \frac{3}{5}v$.
$$|v_1| = \frac{3}{5} \left(5\sqrt{6\mu gd}\right) = 3\sqrt{6\mu gd}$$

After rebounding, the mass falls off the edge from a height $d$. This is a classic horizontal projectile motion problem. The time of flight $t$ depends only on the vertical height:
$$t = \sqrt{\frac{2d}{g}}$$

The horizontal distance $x$ it covers before hitting the ground is simply its horizontal speed multiplied by the time of flight:
$$x = |v_1| t = \left(3\sqrt{6\mu gd}\right) \left(\sqrt{\frac{2d}{g}}\right)$$
$$x = 3\sqrt{12\mu d^2} = 6d\sqrt{3\mu}$$

And there we have it! A beautiful synthesis of collision mechanics, rigid body stability, and projectile motion.

Visualized Solution

System Setup

Elastic Collision

Velocities After Impact

Friction on Block B

Stopping Distance

Toppling Condition

Minimum Velocity $v_{0}$

Rebound Speed

Projectile Motion

The Sigma Insight: Conservation of Linear Momentum

The Setup

The Elastic Collision

The Deceleration

The Toppling Condition

The Rebound

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