Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The rat

Q: Why can't we just use KE = 1/2 mv^2?

You can, but it's a longer route. You would first need to find the ratio of their velocities using p = mv (since p is constant, v is inversely proportional to m). Then substitute that velocity ratio into the KE formula. Using KE = p^2 / 2m directly skips that intermediate step because it already incorporates momentum.

Q: If momentum is the same, shouldn't the energy be the same?

Not at all! Momentum is a measure of 'quantity of motion' (mass times velocity), while kinetic energy is a measure of the 'work' an object can do due to its motion (depends on velocity squared). A light object needs a much higher velocity to have the same momentum as a heavy object. Because velocity is squared in the energy formula, the lighter, faster object ends up with more kinetic energy.

Q: What if the kinetic energies were equal instead of momentum?

If kinetic energies were equal, we would use the same formula KE = p^2 / 2m, but rearrange it to p = sqrt(2m * KE). In that case, momentum would be directly proportional to the square root of the mass. The heavier object would have more momentum.

Q: I got the ratio as 1/2. Where did I go wrong?

You likely assumed kinetic energy is directly proportional to mass. Remember, KE = p^2 / 2m. Since mass is in the denominator, KE is inversely proportional to mass when momentum is constant. So, KE_A / KE_B = m_B / m_A, not m_A / m_B.

Q: Is this a common trick in JEE?

Yes! Problems asking you to compare KE when momentum is constant (or vice versa) are very common. Always remember the relation KE = p^2 / 2m to solve these instantly.

Question

Two solids $A$ and $B$ of mass $1	ext{ kg}$ and $2	ext{ kg}$ respectively are moving with equal linear momentum. The ratio of their kinetic energies $(	ext{KE})_A : (	ext{KE})_B$ will be $\frac{A}{1}$, so the value of $A$ will be ……… .

Sigma Percentile · Accepted Answer

### The Dance of Momentum and Kinetic Energy

Imagine you are standing on a frictionless ice rink. A heavy bowling ball and a light tennis ball are sliding towards you. Surprisingly, they both have the exact same linear momentum. Which one should you be more afraid of?

To answer this, we need to dive into the beautiful relationship between momentum, mass, and kinetic energy. This problem is a classic test of how well you understand this interplay.

### The Master Equation

We know the standard formula for kinetic energy is:
$$ 	ext{KE} = \frac{1}{2}mv^2 $$

And linear momentum is defined as:
$$ p = mv $$

If we multiply and divide the kinetic energy equation by mass $m$, we get a magical transformation:
$$ 	ext{KE} = \frac{1}{2} \frac{m^2 v^2}{m} = \frac{p^2}{2m} $$

This equation is a powerful tool in physics. It tells us exactly how kinetic energy behaves when momentum is held constant.

### Analyzing the Setup

In our problem, we have two solids, $A$ and $B$. 
- Mass of $A$, $m_A = 1	ext{ kg}$
- Mass of $B$, $m_B = 2	ext{ kg}$

The crucial constraint given is that their linear momenta are equal: $p_A = p_B = p$.

Looking at our master equation, $	ext{KE} = \frac{p^2}{2m}$, if $p$ is a constant, the numerator is fixed. This means the kinetic energy is **inversely proportional** to the mass.
$$ 	ext{KE} \propto \frac{1}{m} $$

This is a profound physical insight! It means that for two objects with the same momentum, the lighter object must be moving much faster to compensate for its lack of mass. And because kinetic energy scales with the *square* of velocity, the lighter, faster object ends up packing way more kinetic energy.

### Final Calculation

Let's set up the ratio for our two solids:
$$ \frac{(	ext{KE})_A}{(	ext{KE})_B} = \frac{m_B}{m_A} $$

Substituting the given masses:
$$ \frac{(	ext{KE})_A}{(	ext{KE})_B} = \frac{2}{1} $$

The problem states that this ratio is equal to $\frac{A}{1}$. By simple comparison:
$$ \frac{A}{1} = \frac{2}{1} $$

Therefore, the value of **$A = 2$**.

Going back to our ice rink analogy, the lighter tennis ball has much more kinetic energy than the heavy bowling ball, even though their momenta are the same. It would definitely sting more!

Animated Solution for Physics - Work, Energy, and Power: Two solids $A$ and $B$ of mass $1 kg$ and $2 kg$ respectively are moving with equal linear momentum. The ratio of their kinetic energies $(KE)_{A} : (KE)_{B}$ will be $\frac{A}{1}$ , so the value of $A$ will be ……… .

Enter Numerical Value:

Visualized Solution

Visualizing the System

The Master Formula

Establishing the Proportionality

Setting up the Ratio

Substituting the Values

Final Conclusion

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Dance of Momentum and Kinetic Energy

The Master Equation

Analyzing the Setup

Final Calculation

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