JEE Main 2019, 8 April Shift-II

LEVELJEE Main

Animated Solution for Physics - System of Particles: A body of mass $m_{1}$ moving with an unknown velocity of $v_{1} \hat{i}$ , undergoes a collinear collision with a body of mass $m_{2}$ moving with a velocity $v_{2} \hat{i}$ . After collision, $m_{1}$ and $m_{2}$ move with velocities of $v_{3} \hat{i}$ and $v_{4} \hat{i}$ , respectively. If $m_{2} = 0.5 m_{1}$ and $v_{3} = 0.5 v_{1}$ , then $v_{1}$ is

Q: Why do we ignore the $\hat{i}$ vector in our calculations?

The $\hat{i}$ vector simply indicates that all velocities are directed along the x-axis. Since the motion is strictly one-dimensional (collinear), we can drop the vector notation and just use the scalar magnitudes with appropriate signs.

Q: Why did we substitute $m_2 = m$ and $m_1 = 2m$ instead of just using $m_1$ and $m_2$?

You can absolutely solve it using $m_1$ and $0.5m_1$! However, substituting $m$ and $2m$ removes the decimals and makes the algebraic manipulation slightly cleaner and less prone to silly calculation errors.

Q: What if the second mass was moving in the opposite direction initially?

If $m_2$ was moving towards $m_1$, its initial velocity would be $-v_2 \hat{i}$. You would simply plug in a negative value for $v_2$ in the momentum conservation equation. The fundamental principle remains exactly the same.

Q: Is it possible for $v_1$ to be negative based on the final answer $v_1 = v_4 - v_2$?

Yes! If $v_2$ is greater than $v_4$, then $v_1$ would be negative. Physically, this would mean that the initial velocity of $m_1$ was actually in the $-\hat{i}$ direction. The math naturally takes care of the directions.

Select Answer:

Visualized Solution

$Initial Setup$

$m_{1}$ moving with $v_{1} \hat{i}$
$m_{2}$ moving with $v_{2} \hat{i}$

$Conservation of Momentum$

$p_{initial} = p_{final}$
$m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{3} + m_{2} v_{4}$

$Substituting Given Values$

Given: $m_{2} = 0.5 m_{1} \Rightarrow m_{1} = 2 m_{2}$
Let $m_{2} = m$ , then $m_{1} = 2 m$
Given: $v_{3} = 0.5 v_{1}$

$Applying Substitutions$

$(2 m) v_{1} + (m) v_{2} = (2 m) (0.5 v_{1}) + (m) v_{4}$

$Simplifying the Equation$

$2 m v_{1} + m v_{2} = m v_{1} + m v_{4}$
Divide by $m$ :
$2 v_{1} + v_{2} = v_{1} + v_{4}$

$Solving for v_{1}$

$2 v_{1} - v_{1} = v_{4} - v_{2}$
$v_{1} = v_{4} - v_{2}$

$Beyond Momentum$

Check for kinetic energy loss to determine collision type.
$Δ K = K_{i} - K_{f}$

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The Sigma Insight: Conservation of Linear Momentum

Solution Diagram

Imagine you are observing a straight-line track where two objects are moving. The first object, with mass

m_{1}

, is cruising along with a velocity

v_{1} \hat{i}

. Ahead of it is a second object, mass

m_{2}

, moving with velocity

v_{2} \hat{i}

. They are on a direct collision course!

This is a classic 1D collision problem. The

\hat{i}

simply tells us that all the action is happening along the x-axis.

The Power of Momentum Conservation

When these two bodies collide, they exert forces on each other. However, if we consider both bodies together as a single system, these collision forces are purely internal.

Since there are no external horizontal forces acting on our system, the total linear momentum must remain perfectly conserved. This is our master key to unlocking the problem.

We can write the momentum conservation equation as:

m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{3} + m_{2} v_{4}

Simplifying the Algebra

Now, the problem gives us some fantastic clues to simplify our lives. We are told that

m_{2} = 0.5 m_{1}

To make the algebra cleaner and avoid decimals, let's set

m_{2} = m

. This immediately means

m_{1} = 2 m

We are also given a relationship for the final velocity of the first mass:

v_{3} = 0.5 v_{1}

Let's carefully substitute these relationships into our momentum equation:

(2 m) v_{1} + (m) v_{2} = (2 m) (0.5 v_{1}) + (m) v_{4}

The Final Reveal

Look closely at the equation we just built. Every single term contains the mass

m

. This is a beautiful moment in physics where the specific mass value doesn't matter, only the ratio does! We can safely divide the entire equation by

m

2 v_{1} + v_{2} = 2 (0.5 v_{1}) + v_{4}

Simplifying the right side,

2 \times 0.5

is just

1

2 v_{1} + v_{2} = v_{1} + v_{4}

Our goal is to isolate

v_{1}

. Let's subtract

v_{1}

from both sides:

v_{1} + v_{2} = v_{4}

Finally, moving

v_{2}

to the other side gives us our answer:

v_{1} = v_{4} - v_{2}

And there we have it! By simply trusting the conservation of momentum and carefully substituting our given ratios, we've arrived at the exact expression for

v_{1}

. This perfectly matches option (d).

Common Traps & FAQs

Why do we ignore the

\hat{i}

vector in our calculations?

The

\hat{i}

vector simply indicates that all velocities are directed along the x-axis. Since the motion is strictly one-dimensional (collinear), we can drop the vector notation and just use the scalar magnitudes with appropriate signs.

Can we use the conservation of kinetic energy to solve this?

No, we cannot. The problem does not state that the collision is perfectly elastic. In inelastic collisions, kinetic energy is lost (usually as heat or sound), so equating initial and final kinetic energy would give an incorrect result.

Why did we substitute

m_{2} = m

and

m_{1} = 2 m

instead of just using

m_{1}

and

m_{2}

You can absolutely solve it using

m_{1}

and

0.5 m_{1}

! However, substituting

m

and

2 m

removes the decimals and makes the algebraic manipulation slightly cleaner and less prone to silly calculation errors.

What if the second mass was moving in the opposite direction initially?

m_{2}

was moving towards

m_{1}

, its initial velocity would be

- v_{2} \hat{i}

. You would simply plug in a negative value for

v_{2}

in the momentum conservation equation. The fundamental principle remains exactly the same.

Is it possible for

v_{1}

to be negative based on the final answer

v_{1} = v_{4} - v_{2}

Yes! If

v_{2}

is greater than

v_{4}

, then

v_{1}

would be negative. Physically, this would mean that the initial velocity of

m_{1}

was actually in the

- \hat{i}

direction. The math naturally takes care of the directions.

Select Answer:

Visualized Solution

$Initial Setup$

$Conservation of Momentum$

$Substituting Given Values$

$Applying Substitutions$

$Simplifying the Equation$

$Solving for v_{1}$

$Beyond Momentum$

The Sigma Insight: Conservation of Linear Momentum

The Power of Momentum Conservation

Simplifying the Algebra

The Final Reveal

Similar Questions

Two particles of equal mass $m$ have respective initial velocities $u \hat{i}$ and $u (\frac{i ^ + j ^}{2})$ . They collide completely inelastically. The energy lost in the process is

A body of mass $1$ kg initially at rest, explodes and breaks into three fragments of masses in the ratio $1 : 1 : 3$ . The two pieces of equal mass fly-off perpendicular to each other with a speed of $30$ m/s each. What is the velocity of the heavier fragment ?

A block moving horizontally on a smooth surface with a speed of $40 ms^{- 1}$ splits into two equal parts. If one of the parts moves at $60 ms^{- 1}$ in the same direction, then the fractional change in the kinetic energy will be $x : 4$ , where $x$ is

Select Answer:

Visualized Solution

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for v1​

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

The Sigma Insight: Conservation of Linear Momentum

The Power of Momentum Conservation

Simplifying the Algebra

The Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two particles of equal mass m have respective initial velocities ui^ and u(2i^+j^​​). They collide completely inelastically. The energy lost in the process is

$Initial Setup$

$Conservation of Momentum$

$Substituting Given Values$

$Applying Substitutions$

$Simplifying the Equation$

$Solving for v_{1}$

$Beyond Momentum$

Two particles of equal mass $m$ have respective initial velocities $u \hat{i}$ and $u (\frac{i ^ + j ^}{2})$ . They collide completely inelastically. The energy lost in the process is