Momentum Height Graph for Vertical Motion

❓ Question

A ball is thrown vertically upward from the ground.

The correct momentum-height $(p-h)$ graph is:

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Conservation of mechanical energy
👉 Relation between momentum and height
👉 Motion under gravity.

Main idea:

As height increases:

$$\text{velocity decreases}$$

Hence momentum also decreases.

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🔷 Step 1 — Use Energy Conservation 💯

For vertically upward motion:

$$\frac12 mv^2+mgh=\text{constant}$$

At ground:

$$h=0,\quad v=u$$

So:

$$\frac12 mv^2+mgh=\frac12 mu^2$$

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🔷 Step 2 — Express Velocity in Terms of Height

Rearranging:

$$v^2=u^2-2gh$$

Taking square root:

$$v=\sqrt{u^2-2gh}$$

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🔷 Step 3 — Momentum Relation

Momentum:

$$p=mv$$

So:

$$p=m\sqrt{u^2-2gh}$$

This is not a straight line.

As height increases:

$$p \text{ decreases non-linearly}$$

and finally becomes zero at maximum height.

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🔷 Step 4 — Shape of Graph

At:

$$h=0$$

momentum is maximum.

As:

$$h \uparrow$$

momentum gradually decreases and curve bends toward x-axis.

Hence correct graph is the decreasing curved graph.

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🔷 Step 5 — Correct Option

$$\boxed{(2)}$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ $p-h$ graph ko straight line assume kar lena

❌ Momentum negative lena during upward motion

❌ $v^2=u^2-2gh$relation use na karna

Remember:

$$p\propto\sqrt{u^2-2gh}$$

So graph is curved, not linear.

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✅ Final Answer

$$\boxed{(2)}$$

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