JEE Mains - Collision & Projectile Motion
Question: A ball of mass \(0.2\,\text{kg}\) rests on a vertical post of height \(20\,\text{m}\). A bullet of mass \(0.01\,\text{kg}\), traveling horizontally, hits the center of the ball. After the collision, both travel independently. The ball hits the ground at a horizontal distance of \(30\,\text{m}\) from the foot of the post, and the bullet at a horizontal distance of \(120\,\text{m}\) from the foot of the post. Given \(g = 10\,\text{m/s}^2\), determine the initial velocity of the bullet.
[30-Jan-2023 Shift 1]
Select the correct answer:
Solution
Step 1: Compute the time taken to fall from a height of \(20\,\text{m}\).
Using the equation for free fall,
\[
t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 20}{10}} = \sqrt{4} = 2\,\text{s}.
\]
Step 2: Find the horizontal velocities after the collision.
The horizontal distances traveled during the \(2\,\text{s}\) flight are given by:
For the ball:
\[
v_{\text{ball}} = \frac{30}{2} = 15\,\text{m/s}.
\]
For the bullet:
\[
v_{\text{bullet}} = \frac{120}{2} = 60\,\text{m/s}.
\]
Step 3: Apply conservation of horizontal momentum during the collision.
Initially, the ball is at rest. Let the initial speed of the bullet be \(u\). Then, conservation of momentum gives:
\[
m_{\text{bullet}} \cdot u = m_{\text{ball}} \cdot v_{\text{ball}} + m_{\text{bullet}} \cdot v_{\text{bullet}}.
\]
Plug in the values:
\[
0.01\,u = (0.2 \times 15) + (0.01 \times 60) = 3 + 0.6 = 3.6.
\]
Therefore,
\[
u = \frac{3.6}{0.01} = 360\,\text{m/s}.
\]
Hence, the initial velocity of the bullet is \(360\,\text{m/s}\).
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