Question

A bullet of mass $m_1=20gm$ fired horizontally with a velocity of $v=200m/s$ hits a wooden block of mass $m_2=500$ gm (take $g=9.8m/s^2$) 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
• B. 10.06 m
• C. 20.12 m
• D. 100 m

Answer 1

The correct option is B 10.06 m

Initial momentum of bullet and block
= final momentum of bullet and block

$=\frac{20}{1000}\times 200+0=\frac{(500+20)}{1000}\times V$

$\therefore v=7.69m/se{c}^2$

Now apply work energy principle

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

$\frac{1}{2}m{v}^2-0=\mu \times mgx$

$\frac{1}{2}\times \left(\frac{520}{1000}\right)\times {7.62}^2=0.3\times \frac{520}{1000}\times 9.81\times x$

$\therefore x=10.046m\approx 10.05m$