Question

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

• A. $0.16$
• B. $0.8$
• C. $0.32$
• D. $0.24$

Answer 1

The correct option is A $0.16$

Mass of the bullet $=0.2kg=m_1$

Mass of the block $=10kg=m_2$

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

Initial velocity of block$=0m/s=v_2$

Final velocity of bullet$=v_1^{\prime }=100m/s$

Final velocity of block$=v_2^{\prime }$

From the conservation of momentum:

$m_1{v}_1+m_2{v}_2=m_1{v}_1^{\prime }+m_2{v}_2^{\prime }$

$0.02\times 500+10\times 0=0.02\times 100+10v_2^{\prime }$

$10=2+10v_2^{\prime }$

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

After moving a distance of 0.2 the block stops.Therefore change in kinetic energy =work done by friction

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

$\mu =0.16$