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kiruha [24]
3 years ago
6

If a bullet loses 1/nth of its velocity while passing through a plank,then no of planks required to stop the bullet is?

Physics
1 answer:
Novay_Z [31]3 years ago
8 0
Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing <span><span>1/nth</span><span>1/nth</span></span><span> of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).

</span><span>With this assumption, the energy loss becomes
</span><span>
ΔE = <span>1/2 </span>m<span>v2 </span>− <span>1/2 </span>m <span><span>(<span>v−<span>v/n</span></span>) </span><span>2
</span></span></span>
and the number of planks <span>NN</span><span> becomes
</span>
N = <span><span><span>1/2</span>m<span>v2 /</span></span><span>ΔE </span></span>= <span><span>n2/ </span><span>2n−1
</span></span>
Otherwise, if you assume that the bullet loses <span><span>1/<span>nth</span></span><span>1/<span>nth</span></span></span><span> of its velocity per plank, then the answer is </span><span><span>N=∞</span></span><span><span>

</span>
</span>
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The unknown mass of the bullet is  m1=3.751x10^{-3} kg

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According to Newton's laws of motion, when a net external force acts on a body of mass  <u><em>m</em></u> , it results in change in momentum of the body and is given by:

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m_{1}*v_{1}*cos(180)+m_{2}*v_{2}*cos(60)+m_{3}*v_{3}*cos(-60)=0\\m_{1}*605\frac{m}{s}*cos(180)+5.79x10^{-3}kg *392\frac{m}{s}*cos(60)+5.79x10^{-3}kg*392\frac{m}{s}*cos(-60)=0\\

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