An object of mass m1 collides with another object of mass m2, which is at rest. After the collision,

Q: Why did we assume the collision is perfectly elastic?

The problem states they move with equal speeds in opposite directions but doesn't give enough information to solve for the mass ratio if energy is lost. In standard JEE problems, if a unique ratio is expected without specifying the collision type, it implies an ideal, perfectly elastic collision ($e=1$).

Q: Why is the final velocity of $m_1$ taken as $-v$?

Velocity is a vector. If we take the initial direction of $m_1$ (towards the right) as positive, and it bounces back (moves in the opposite direction) after the collision, its velocity must be negative.

Q: Can we solve this using the conservation of kinetic energy instead of the coefficient of restitution?

Absolutely! For an elastic collision, kinetic energy is conserved. You can write $\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v^2 + \frac{1}{2}m_2 v^2$. Substituting $u_1$ from the momentum equation will yield the exact same result, though the algebra might be slightly longer.

Q: What if the objects moved in the same direction after the collision?

If they moved in the same direction, the final velocity of $m_1$ would be $+v$. The momentum equation would be $m_1 u_1 = (m_1 + m_2)v$. However, for an elastic collision, a lighter object striking a heavier stationary object will always bounce back.

Q: Does the final ratio depend on the initial speed $u_1$?

No, it doesn't! As you can see in the derivation, the velocity terms completely cancel out. The ratio of the masses is an intrinsic property of this specific collision outcome, independent of how fast the first mass was initially moving.

Question

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

Options:

(A) 3 : 1

(B) 2 : 1

(C) 1 : 2

(D) 1 : 1

Sigma Percentile · Accepted Answer

### Setting the Stage: The Collision

Imagine a classic physics scenario: a mass $m_1$ is cruising along with an initial speed $u_1$, heading straight for a stationary mass $m_2$.

Suddenly, they collide! The problem tells us that after this impact, the two masses bounce back, moving in exactly opposite directions but with the exact same speed, $v$.

Our goal is to find the ratio of their masses, $m_2 : m_1$. To do this, we need to rely on the fundamental laws of mechanics.

### The Master Equation: Conservation of Momentum

Since there are no external forces acting on our two-block system, the total linear momentum must be strictly conserved. This means the initial momentum of the system will perfectly equal the final momentum.

Let's write down the raw values. Initially, only $m_1$ is moving, so the initial momentum is $m_1 u_1$.

Finally, $m_1$ moves backwards with speed $v$, giving it a velocity of $-v$, while $m_2$ moves forwards with a velocity of $+v$. Equating the initial and final states, we get:

$$m_1 u_1 + m_2 (0) = m_1 (-v) + m_2 (v)$$

Let's simplify this equation by grouping the terms with $v$. We find that:

$$m_1 u_1 = (m_2 - m_1)v$$

Let's call this our master momentum equation. It relates the initial speed to the final speed, but we still have two unknowns ($u_1$ and $v$). We need another equation!

### The Hidden Assumption: Perfect Elasticity

There is a catch here. The problem doesn't explicitly state the nature of the collision. Is kinetic energy conserved? Is some energy lost as heat or sound?

In such JEE problems, to find a unique ratio without additional data, we must assume a **perfectly elastic collision**. For an elastic collision, the coefficient of restitution, $e$, is exactly equal to $1$.

The coefficient of restitution is the ratio of the relative velocity of separation to the relative velocity of approach.

$$e = \frac{v_2 - v_1}{u_1 - u_2}$$

The separation velocity is $v - (-v)$, and the approach velocity is simply $u_1 - 0$. Equating this to $1$, we get:

$$1 = \frac{v - (-v)}{u_1 - 0}$$

Watch out for the minus sign here! The numerator becomes $2v$. This reveals a beautiful relationship:

$$u_1 = 2v$$

The initial speed $u_1$ is exactly twice the final speed $v$.

### The Final Reveal: Solving for the Ratio

Now, remember that master momentum equation we derived earlier? Let's bring that back and carefully substitute $u_1 = 2v$ into it.

$$m_1 (2v) = (m_2 - m_1)v$$

Notice how elegantly the $v$ cancels out from both sides. We are left with:

$$2m_1 = m_2 - m_1$$

Shifting $m_1$ to the left side gives us:

$$3m_1 = m_2$$

Therefore, the ratio of $m_2$ to $m_1$ is exactly:

$$\frac{m_2}{m_1} = \frac{3}{1}$$

A brilliant and clean result! The stationary mass must be exactly three times heavier than the incoming mass for them to bounce apart with equal speeds.

Animated Solution for Physics - System of Particles: 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

Select Answer:

Visualized Solution

$Visualizing the Collision$

$Conservation of Linear Momentum$

$Substituting Values$

$Simplifying Momentum Equation$

$Coefficient of Restitution$

$Applying Restitution$

$Relation between u_{1} and v$

$Substituting u_{1} in Eq (i)$

$Solving for Mass Ratio$

$Conclusion$

The Sigma Insight: Head-on Collision

Setting the Stage

The Master Equation

The Hidden Assumption

The Final Reveal

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Collision

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Linear Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying Momentum Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coefficient of Restitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Restitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relation between u1​ and v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting u1​ in Eq (i)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Mass Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion