Question

A bullet of mass $20g$ travelling horizontally with a speed of $500m/s$ passes through a wooden block of mass $10.0Kg$ initially at rest to a level surface. The bullet emerges with a speed of $100m/s$ and the block slides $20cm$ on the surface before coming to rest. Find the friction coefficient between the block and the surface in the figure. $(g=10m/s^2)$

Answer 1

Given that,

Mass of bullet $m=0.02kg$

Initial velocity of bullet $v_1=500m/s$

Mass of block $M=10kg$

Initial velocity $v_2=0$

Final velocity ${v^{\prime }}_1=100m/s$

Now, the final velocity of block when bullet comes out

If block velocity $=v_2^{\prime }$

$m{v}_1+M{v}_2=m{v}_1+M{v}_2^{\prime }$

$0.02\times 500=0.02\times 100+10v_2^{\prime }$

$v_2^{\prime }=\frac{10-2}{10}$

$v_2^{\prime }=0.8m/s$

Now, after moving 0.2 m

Change in kinetic energy = work done

$0-\frac{1}{2}\times 10\times {\left(0.8\right)}^2=-\mu \times 10\times 10\times 0.2$

$\mu =0.16$

Hence, the friction coefficient is $0.16$