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A bullet of mass m1=20gm fired horizontally with a velocity of v=200m/s hits a wooden block of mass m2=500 gm (take g=9.8m/s2) resting on a horizontal plane as shown in the figure and the bullet remains embedded in the block after the impact. If the coefficient of friction between the surface in contact remains constant at 0.3, the distance the block will move before coming to rest is


A
5.03 m
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B
10.06 m
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C
20.12 m
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D
100 m
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Solution

The correct option is B 10.06 m
Initial momentum of bullet and block
= final momentum of bullet and block

=201000×200+0=(500+20)1000×V

v=7.69 m/sec2

Now apply work energy principle

(final kinetic energy) - (initial kinetic energy)
= network done

12mv20=μ×mgx

12×(5201000)×7.622=0.3×5201000×9.81×x

x=10.046 m10.05m

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