A block of mass \(1.9\, kg\) is at rest at the edge of a table, of height \(1\, m\). A bullet of mass \(0.1\, kg\) collides with the block and sticks to it. If the velocity of the bullet is \(20\, m / s\) in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is \(....J\) [Take \(g =10\, m / s ^{2}\). Assume there is no rotational motion and loss of energy after the collision is negligable. \(]\)

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Verified Solution

Answer & Solution

Correct Answer

(A) \(21\)

Step-by-step Solution

Detailed explanation

\(p _{ i }= p _{ f } \Rightarrow 0.1 \times 20=2 v\)

\(\therefore v =1 m / s\)

\(KF _{ f }= mgh +\frac{1}{2} mv ^{2}=20+1=21\)

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