Laws of Motion

JEE Main level, single correct, medium

184 of 210

Question 184 of 210

A bullet of mass 4 g4 \mathrm{~g} is fired horizontally with a speed of 300 m/s300 \mathrm{~m} / \mathrm{s} into 0.8 kg0.8 \mathrm{~kg} block of wood at rest on a table. If the coefficient of friction between the block and the table is 0.30.3, how far will the block slide approximately?

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Answer: (B) 0.379 m0.379 \mathrm{~m}

Given, m1=4 g,u1=300 m/s\mathrm{m}_1=4 \mathrm{~g}, \mathrm{u}_1=300 \mathrm{~m} / \mathrm{s} m2=0.8 kg=800 g,u2=0 m/s\mathrm{m}_2=0.8 \mathrm{~kg}=800 \mathrm{~g}, \mathrm{u}_2=0 \mathrm{~m} / \mathrm{s} From law of conservation of momentum, m1u1+m2u2=m1v1+m2v2\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2 Let the velocity of combined system =vm/s=\mathrm{v} \mathrm{m} / \mathrm{s} then, 4×300+800×0=(800+4)×v4 \times 300+800 \times 0=(800+4) \times v v=1200804=1.49 m/s\mathrm{v}=\frac{1200}{804}=1.49 \mathrm{~m} / \mathrm{s}
Now, μ=0.3\mu=0.3 (given)
a=μga=0.3×10( take g=10 m/s2)=3 m/s2\begin{aligned} &\mathrm{a}=\mu \mathrm{g} \\ &\mathrm{a}=0.3 \times 10\left(\text { take } \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right) \\ &=3 \mathrm{~m} / \mathrm{s}^2 \end{aligned}
then, from v2=u2+2as\mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as}
(1.49)2=0+2×3×s(1.49)^2=0+2 \times 3 \times s

s=(1.492)6;s=2.226=0.379 m\mathrm{s}=\frac{\left(1.49^2\right)}{6} ; \mathrm{s}=\frac{2.22}{6}=0.379 \mathrm{~m}

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