Question

A bomb explodes in air when it has a horizontal speed of v. It breaks into two identical pieces of equal mass. If one goes vertically up at a speed of 4v, the velocity of other immediately after the explosion is ________.

• A. $4\hat{j}$
• B. $-v\hat{i}$
• C. $2v\hat{i}-4v\hat{j}$
• D. $2v\hat{i}+4v\hat{j}$

Hint:

Internal explosions conserve total momentum in all vector directions.

Answer 1

The Correct Option is C

Step 1: Concept
Conservation of Linear Momentum: $\vec{P}_{initial} = \vec{P}_{final}$.
Step 2: Analysis
Initial momentum = $(2m)v\hat{i}$ (assuming total mass is $2m$). Piece 1 momentum = $m(4v\hat{j})$. Piece 2 velocity = $\vec{v}_2$.
Step 3: Calculation
$2mv\hat{i} = m(4v\hat{j}) + m\vec{v}_2$. Divide by $m$: $2v\hat{i} = 4v\hat{j} + \vec{v}_2$. $\vec{v}_2 = 2v\hat{i} - 4v\hat{j}$.
Step 4: Conclusion
Hence, the velocity of the other piece is $2v\hat{i} - 4v\hat{j}$.
Final Answer:(C)