1.

Q23. A Bullet flying with a velocity of50ms-1 hits 'a block of wood andpenetrates through a distance of 0.2m before coming to rest. The Mass ofthe bullet is 0.03kg. Calculate theresistance offered by the block ofwood.​

Answer»

ANSWER:

Initial velocity, u=50 ms

−1

Distance TRAVELLED, s = 0.2

FINAL velocity v = 0

Mass of the bullet, m = 0.03 KG

From the equation, v

2

=u

2

+2as

0=50

2

+2×a×0.2

a=−

0.4

2500

=−6250 ms

−2

∴ The resistance OFFERED by the wood is

F=ma=0.03×6250=187.5N.

Explanation:

hope this helps you :)



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