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  • A bullet of mass moving horizontally with speed <img alt="400 \mathrm{~ms}^{-1}" src=

A bullet of mass 0.1 \mathrm{~kg} moving horizontally with speed 400 \mathrm{~ms}^{-1} hits a wooden block of mass 3.9 \mathrm{~kg} kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 \mathrm{~m} before coming to rest. The coefficient of friction between the block and the surface is______.
(Given g=10 \mathrm{~m} / \mathrm{s}^{2})

Option: 1

0.90


Option: 2

0.65


Option: 3

0.25


Option: 4

0.50


Answers (1)

best_answer



Apply momentum conservation just before and just after the collision

0.1 \times 400=(3.9+.1) u^{\prime}

\Rightarrow \mathrm{u}=10 \mathrm{~m} / \mathrm{s}

\Delta \mathrm{KE}=\mathrm{w}_{\text {all FORCE }}

\because \mathrm{f}=\mu \mathrm{mg} (kinetic friction)

\Rightarrow 0-\frac{1}{2}(4)(10)^{2}=-\mu(4) g \times 20

\Rightarrow \mu=0.25

Posted by

Kuldeep Maurya

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