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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. |
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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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