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JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 15)

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 [Take g = 10 m/s2 . Assume there is no rotational motion and loss of energy after the collision is negligable.]
23 J
21 J
20 J
19 J

Explanation

JEE Main 2020 (Online) 3rd September Evening Slot Physics - Center of Mass and Collision Question 69 English Explanation

Let velocity of block after collision = v

Using momentum conservation,

pi = pf

0.1×20 = (1.9 + 0.1)V

$$ \Rightarrow $$ 2 = 2 V

$$ \Rightarrow $$ V = 1 m/sec

Kinetic energy just before striking the floor

= $${1 \over 2}m{v^2} + mgh$$

= $${1 \over 2} \times 2{\left( 1 \right)^2} + 2 \times 10 \times 1$$

= 21 J

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