Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These hav

Q: Why can't we just conserve kinetic energy for the whole process?

The second collision is completely inelastic. In inelastic collisions, kinetic energy is not conserved; some of it is lost as heat or deformation energy. We can only conserve kinetic energy during the first (elastic) collision.

Q: Why did we ignore block C during the first collision?

The collisions happen sequentially. Block A hits block B first. During this infinitesimally short time, block C is far away and unaffected. We analyze the system chronologically.

Q: Is the total momentum of the system conserved throughout?

Yes! Since there are no external horizontal forces (frictionless surface), the total momentum of the $A+B+C$ system is conserved from the very beginning to the very end.

Q: What happens to block A after the first collision?

Block A bounces back! Using the formula $v_A = \frac{m_A - m_B}{m_A + m_B}u_A$, we get $v_A = \frac{m - 2m}{3m}(9) = -3 \text{ m/s}$. The negative sign indicates it moves to the left.

Q: Could we solve this without finding $v_B$?

Absolutely! Using the total system momentum conservation: $p_{\text{initial}} = p_{\text{final}} \implies m(9) = m(-3) + (2m+m)v_C$. Solving this gives $v_C = 4 \text{ m/s}$ directly. This is a great shortcut!

Question

Three objects $A, B$ and $C$ are kept in a straight line on a frictionless horizontal surface. These have masses $m, 2m$ and $m$, respectively. The object $A$ moves towards $B$ with a speed $9 	ext{ ms}^{-1}$ and makes an elastic collision with it. Thereafter, $B$ makes completely inelastic collision with $C$. All motions occur on the same straight line. Find the final speed (in $	ext{ms}^{-1}$) of the object $C$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a frictionless horizontal surface where three blocks, $A$, $B$, and $C$, are placed in a straight line. The masses are $m_A = m$, $m_B = 2m$, and $m_C = m$.

Block $A$ is fired towards block $B$ with an initial speed of $u_A = 9 	ext{ m/s}$. The problem states that two distinct collisions will occur sequentially. First, an **elastic collision** between $A$ and $B$. Then, a **completely inelastic collision** between $B$ and $C$.

Our goal is to find the final velocity of block $C$ after all the dust settles.

### Phase 1: The Elastic Strike

When block $A$ collides with the stationary block $B$, the collision is perfectly elastic. This means both momentum and kinetic energy are conserved.

Instead of writing out both conservation equations and solving the quadratic system, we can use the standard derived formula for the velocity of a stationary target after a 1D elastic collision:

$$v_B = \frac{2m_A}{m_A + m_B} u_A$$

Let's substitute our known values into this powerful tool:

$$v_B = \frac{2(m)}{m + 2m} (9)$$

Simplifying the mass ratio, we get:

$$v_B = \frac{2m}{3m} 	imes 9 = \frac{2}{3} 	imes 9 = 6 	ext{ m/s}$$

So, after the first impact, block $B$ is launched forward with a speed of $6 	ext{ m/s}$.

### Phase 2: The Inelastic Embrace

Now, block $B$ is hurtling towards the stationary block $C$ at $6 	ext{ m/s}$. This second collision is **completely inelastic**.

In a completely inelastic collision, the objects stick together upon impact and move as a single combined mass. While kinetic energy is lost to deformation and heat, **linear momentum is strictly conserved**.

Let's set up the momentum conservation equation for the $B-C$ system. The initial momentum is just the momentum of block $B$, and the final momentum is the momentum of the combined mass moving at a final velocity $v_f$:

$$p_{	ext{initial}} = p_{	ext{final}}$$

$$m_B v_B + m_C (0) = (m_B + m_C) v_f$$

Substituting the masses and the velocity of $B$ we just found:

$$(2m)(6) = (2m + m) v_f$$

$$12m = 3m \cdot v_f$$

Dividing both sides by $3m$, we find the final velocity:

$$v_f = \frac{12m}{3m} = 4 	ext{ m/s}$$

Since block $C$ is now stuck to block $B$, its final speed is exactly $4 	ext{ m/s}$.

### The Ninja Technique: System Momentum Conservation

While the step-by-step chronological method is fantastic for building physical intuition, there is a faster, more elegant way to solve this problem—a true "ninja technique" for competitive exams.

Notice that the entire surface is frictionless. This means there are **zero external horizontal forces** acting on the $A+B+C$ system. Therefore, the total linear momentum of the entire three-block system must remain constant from the very beginning to the very end!

Let's calculate the initial momentum of the entire system before any collisions happen:

$$p_{	ext{initial}} = m_A u_A + m_B (0) + m_C (0) = m(9) = 9m$$

Now, what does the system look like at the very end? Block $A$ has bounced back after the first collision, and blocks $B$ and $C$ are stuck together moving at $v_C$. We need the final velocity of $A$, let's call it $v_A'$. Using the elastic collision formula for the projectile:

$$v_A' = \frac{m_A - m_B}{m_A + m_B} u_A = \frac{m - 2m}{m + 2m} (9) = \frac{-m}{3m} (9) = -3 	ext{ m/s}$$

Now, let's write the final momentum of the entire system:

$$p_{	ext{final}} = m_A v_A' + (m_B + m_C) v_C$$

$$p_{	ext{final}} = m(-3) + (2m + m) v_C = -3m + 3m v_C$$

Equating the initial and final total momentum:

$$9m = -3m + 3m v_C$$

$$12m = 3m v_C$$

$$v_C = 4 	ext{ m/s}$$

Both methods yield the exact same result! The chronological method is safer, but the system momentum method is a beautiful demonstration of the deep conservation laws governing our universe.

Enter Numerical Value:

Visualized Solution

The Setup

Elastic Collision: A $\to B$

Substituting Values for $v_{B}$

Calculating $v_{B}$

Inelastic Collision: $B \to C$

Momentum Conservation Setup

Calculating Final Velocity

Final Answer

The Sigma Insight: Head-on Collision

Analyzing the Setup

Phase 1

Phase 2

The Ninja Technique

Similar Questions

A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one-fourth of its original speed. What is the mass of the second body?

A body of mass $2 kg$ moving with a speed of $4 m/s$ makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its initial speed. The speed of the two body centre of mass is $\frac{x}{10} m/s$ . Then, the value of $x$ is ......... .

An object of mass $m_{1}$ collides with another object of mass $m_{2}$ , which is at rest. After the collision, the objects move with equal speeds in opposite direction. The ratio of the masses $m_{2} : m_{1}$ is

A block of mass $0.50 kg$ is moving with a speed of $2.00 ms^{- 1}$ on a smooth surface. It strikes another mass of $1.00 kg$ and then they move together as a single body. The energy loss during the collision is

A point mass of $1 kg$ collides elastically with a stationary point mass of $5 kg$ . After their collision, the $1 kg$ mass reverses its direction and moves with a speed of $2 ms^{- 1}$ . Which of the following statement(s) is/are correct for the system of these two masses?