First slide

Conservation of mechanical energy

Question

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

Easy

A

20J
B

19J
C

23J
D

21J

Solution

$C o n s e r v i n g m o m e n t u m, (1.9) (0) + (0.1) (20) = (1.9 + 0.1) v \Rightarrow v = 1 m / s$
Then conserving kinetic energy when it hits the ground $= I n i t i a l k i n e t i c e n e r g y + g a i n i n p o t e n t i a l e n e r g y = \frac{1}{2} 2 {(1)}^{2} + 2 (10) 1$
$K . E_{f i n a l} = 21 J$

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