A block of mass 0.50 kg is moving with a speed of 2.00 m/s on a smooth surface. It strikes another m

Q: Why isn't kinetic energy conserved in this collision?

In a perfectly inelastic collision, the objects stick together. The process of them sticking together requires energy to deform the materials, and some energy is also lost as heat and sound. Therefore, macroscopic kinetic energy is not conserved.

Q: Why did we assume the second block was at rest?

The problem states that the first block 'strikes another mass of 1.00 kg'. When a problem doesn't specify the velocity of the target object, it is standard convention in physics problems to assume the target is stationary ($v_2 = 0$).

Q: Is there a shortcut formula to find the energy loss directly?

Yes! For a perfectly inelastic collision, the loss in kinetic energy can be directly calculated using the reduced mass formula: $\Delta K = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (v_1 - v_2)^2$. This is a great time-saver for competitive exams.

Q: What if the surface wasn't smooth?

If the surface had friction, it would act as an external force. However, collision forces are usually so large and act over such a tiny time interval that we can still apply momentum conservation just before and just after the impact (impulse approximation). Friction would only affect the motion before and after the collision phase.

Q: Why is momentum always conserved but not kinetic energy?

Momentum is a vector quantity and is conserved as long as the net external force on the system is zero (Newton's Third Law ensures internal forces cancel out). Kinetic energy is a scalar and can be transformed into other non-mechanical forms of energy (like heat or internal potential energy) during the interaction.

Question

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

Options:

(A) $0.16	ext{ J}$

(B) $1.00	ext{ J}$

(C) $0.67	ext{ J}$

(D) $0.34	ext{ J}$

Sigma Percentile · Accepted Answer

### The Anatomy of a Perfectly Inelastic Collision

Imagine you are observing a physics experiment on a perfectly smooth, frictionless track. A block of mass $m_1 = 0.50	ext{ kg}$ is gliding effortlessly at a speed of $v_1 = 2.00	ext{ ms}^{-1}$. Lying in wait directly in its path is a second block, twice as massive at $m_2 = 1.00	ext{ kg}$, completely at rest ($v_2 = 0$).

When the first block strikes the second, they don't bounce off each other. Instead, they stick together and move forward as a single, combined entity. This is the hallmark of a **perfectly inelastic collision**.

### The Master Equation: Conservation of Momentum

Even though the blocks stick together and deform, the universe demands that one fundamental quantity remains constant: **Linear Momentum**. Because the surface is smooth, there are no external horizontal forces acting on our two-block system. Therefore, the total momentum before the crash must exactly equal the total momentum after the crash.

Mathematically, we express this as:
$$p_{	ext{initial}} = p_{	ext{final}}$$
$$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v$$

Let's substitute our known values into this master equation:
$$0.50 	imes 2 + 1.00 	imes 0 = (0.50 + 1.00) v$$
$$1 = 1.5 v$$

Solving for the final velocity $v$, we get:
$$v = \frac{1}{1.5} = \frac{2}{3}	ext{ ms}^{-1}$$

### The Missing Energy

While momentum is strictly conserved, kinetic energy is a different story. In a perfectly inelastic collision, the maximum possible amount of kinetic energy is lost—converted into heat, sound, and the energy required to permanently deform the objects so they stick together.

To find this energy loss, we must calculate the kinetic energy of the system before and after the impact.

**Initial Kinetic Energy ($K_i$):**
Only the first block is moving initially, so the total initial kinetic energy is:
$$K_i = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} (0.50) (2)^2 = 1	ext{ J}$$

**Final Kinetic Energy ($K_f$):**
After the collision, the combined mass $(m_1 + m_2 = 1.5	ext{ kg})$ moves at velocity $v = \frac{2}{3}	ext{ ms}^{-1}$:
$$K_f = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (1.5) \left(\frac{2}{3}ight)^2$$
$$K_f = \frac{1}{2} 	imes \frac{3}{2} 	imes \frac{4}{9} = \frac{1}{3}	ext{ J}$$

### The Final Calculation

The energy lost during the collision is simply the difference between what we started with and what we ended up with:
$$\Delta K = K_i - K_f$$
$$\Delta K = 1 - \frac{1}{3} = \frac{2}{3}	ext{ J}$$

Converting this fraction to a decimal, we get:
$$\Delta K \approx 0.67	ext{ J}$$

**Pro-Tip for JEE Aspirants:** 
There is a direct, highly efficient formula for calculating the loss of kinetic energy in a perfectly inelastic collision. It saves precious time during exams:
$$\Delta K = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (v_1 - v_2)^2$$

Try plugging the values into this formula yourself! You will arrive at the exact same result of $0.67	ext{ J}$ in a fraction of the time.

Animated Solution for Physics - System of Particles: 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

Select Answer:

Visualized Solution

Visualizing the Collision

Conservation of Momentum

Momentum Equation

Substituting Values

Calculating Final Velocity

Energy Loss Formula

Initial Kinetic Energy

Calculating Initial KE

Final Kinetic Energy

Calculating Final KE

Final Energy Loss

The Sigma Insight: Head-on Collision

The Anatomy of a Perfectly Inelastic Collision

The Master Equation

The Missing Energy

The Final Calculation

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