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QJEEM20Ph7496: JEE Main 2020 Physics - Class 11 JEEM20Ph7496 Solution: A block of mass 1.9 kg is at rest at the edge of a table, of…
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.]
Multiple Choice Options
- A 23 J
- B 21 J
- C 20 J
- D 19 J
Step-by-Step Verified Solution
Let velocity of block after collision = v
Using momentum conservation,
pi = pf
0.1×20 = (1.9 + 0.1)V
2 = 2 V
V = 1 m/sec
Kinetic energy just before striking the floor
=
=
= 21 J