Two masses of 1 g and 4 g are moving with equal kinetic energies. The ratio of the magnitudes of the

Q: Wait, why can't we just use $p = mv$ directly?

You absolutely can, but you don't know their velocities! You only know that their kinetic energies are equal. Using the relation $p = \sqrt{2mK}$ bypasses the need to calculate the individual velocities first, saving you time and algebraic steps.

Q: Do we need to convert the mass from grams to kilograms before solving?

In this specific problem, no. Since we are calculating a ratio $\frac{m_1}{m_2}$, the units cancel out completely. Whether you use grams or convert them to kilograms, the ratio $\frac{1}{4}$ remains exactly the same.

Q: If the second mass is heavier, why is its momentum only twice as much and not four times?

Because it is moving slower! Kinetic energy is $K = \frac{1}{2}mv^2$. For a heavy object and a light object to have the same kinetic energy, the heavy object must have a smaller velocity to compensate for its larger mass. This smaller velocity keeps its momentum from scaling linearly with its mass.

Q: What if the problem said their momenta were equal instead of their kinetic energies?

Great question! If their momenta were equal ($p_1 = p_2$), then using the formula $K = \frac{p^2}{2m}$, we can see that kinetic energy is inversely proportional to mass. In that scenario, the lighter object would actually have more kinetic energy!

Q: Why do we only take the positive square root when finding momentum?

The question specifically asks for the ratio of the *magnitudes* of their momenta. Magnitudes of physical vectors are always positive, so we only consider the positive principal square root when evaluating $\sqrt{\frac{1}{4}}$.

Question

Two masses of $1	ext{ g}$ and $4	ext{ g}$ are moving with equal kinetic energies. The ratio of the magnitudes of their momenta is

Options:

(A) $4 : 1$

(B) $\sqrt{2} : 1$

(C) $1 : 2$

(D) $1 : 16$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine two blocks sliding across a perfectly frictionless floor. One is a tiny block with a mass of $m_1 = 1	ext{ g}$, and the other is a heavier block with a mass of $m_2 = 4	ext{ g}$.

Despite their drastic difference in mass, the problem gives us a fascinating constraint: both blocks possess the exact same kinetic energy. That is, $K_1 = K_2 = K$.

Because the second block is four times heavier, it must be moving significantly slower than the first block to maintain this equality in kinetic energy. But the question doesn't ask for their velocities; it asks for the ratio of the magnitudes of their momenta. To find this, we need a mathematical bridge that directly connects kinetic energy, mass, and momentum.

### The Master Equation

We know the standard formulas for kinetic energy and momentum:
$$K = \frac{1}{2}mv^2$$
$$p = mv$$

While we could solve for velocity $v$ in terms of $K$ and substitute it into the momentum equation, there is a much more elegant way. Let's manipulate the kinetic energy equation by multiplying and dividing the right side by the mass $m$:
$$K = \frac{m \cdot mv^2}{2m} = \frac{(mv)^2}{2m}$$

Since $p = mv$, we can substitute $p$ into the numerator:
$$K = \frac{p^2}{2m}$$

This is our master equation! It beautifully links kinetic energy and momentum without needing to know the velocity. Since we want to compare their momenta, let's isolate $p$:
$$p^2 = 2mK$$
$$p = \sqrt{2mK}$$

### Final Calculation

Now, let's apply this master equation to both of our blocks. For the first block, its momentum is:
$$p_1 = \sqrt{2m_1K}$$

For the second block, its momentum is:
$$p_2 = \sqrt{2m_2K}$$

To find the ratio of their momenta, we simply divide $p_1$ by $p_2$:
$$\frac{p_1}{p_2} = \frac{\sqrt{2m_1K}}{\sqrt{2m_2K}}$$

Because the problem states that both blocks have the exact same kinetic energy $K$, the terms $2$ and $K$ are identical in both the numerator and the denominator. We can cleanly cancel them out:
$$\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$$

This elegant result tells us that when kinetic energies are equal, the ratio of momenta is simply the square root of the ratio of their masses. Now, we just plug in the given mass values, $m_1 = 1	ext{ g}$ and $m_2 = 4	ext{ g}$:
$$\frac{p_1}{p_2} = \sqrt{\frac{1}{4}}$$

Evaluating the square root gives us our final answer:
$$\frac{p_1}{p_2} = \frac{1}{2}$$

Therefore, the ratio of the magnitudes of their momenta is **$1 : 2$**. Even though the second block is four times heavier, its momentum is only twice as much because it is moving slower. The correct option is (c).

Animated Solution for Physics - Work, Energy, and Power: Two masses of $1 g$ and $4 g$ are moving with equal kinetic energies. The ratio of the magnitudes of their momenta is

Select Answer:

Visualized Solution

The Physical Setup

Relating $K$ and $p$

$p = 2 m K$

Setting up the Ratio

$\frac{p _{1}}{p _{2}} = \frac{m _{1}}{m _{2}}$

Substituting Mass Values

$\frac{p _{1}}{p _{2}} = \frac{1}{2}$

The Sigma Insight: Kinetic Energy, Potential Energy and Power

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

Two particles having masses $4 g$ and $16 g$ respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is $n : 2$ . The value of $n$ will be ............ .

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 ……… .

If the kinetic energy of a moving body becomes four times of its initial kinetic energy, then the percentage change in its momentum will be

A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_{c}$ is varying with time $t$ as $a_{c} = k^{2} r t^{2}$ , where $k$ is a constant. The power delivered to the particle by the force acting on it is

A particle of mass $0.2 kg$ is moving in one dimension under a force that delivers a constant power $0.5 W$ to the particle. If the initial speed (in $ms^{- 1}$ ) of the particle is zero, the speed (in $ms^{- 1}$ ) after $5 s$ is

A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = - \frac{k}{2 r ^{2}}$ . Its total energy is

A body of mass $m$ is accelerated uniformly from rest to a speed $v$ in a time $T$ . The instantaneous power delivered to the body as a function of time, is given by

A block of mass $2 kg$ is free to move along the $x$ -axis. It is at rest and from $t = 0$ onwards it is subjected to a time-dependent force $F (t)$ in the $x$ -direction. The force $F (t)$ varies with $t$ as shown in the figure. The kinetic energy of the block after $4.5 s$ is

A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to

Animated Solution for Physics - Work, Energy, and Power: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two masses of 1 g and 4 g are moving with equal kinetic energies. The ratio of the magnitudes of their momenta is