Question

A $4g$ bullet is fired horizontally with a speed of $300m/s$ into $0.8kg$ block of wood, which is at rest on a table. If the coefficient of friction between the block and the table is $0.3$, how far will the block slide approximately?

• A. 0.20 m
• B. 0.375 m
• C. 0.565 m
• D. 0.750 m

Answer 1

Answer: (B)

0.375 m

first apply momentum conservation on the bullet and block

$m_1{v}_1=m_2{v}_2$

$m_1=massofbullet{v}_1=velocityofbullet{m}_2=massofblock{v}_2=velocityofblock$

momentum conservation.

$\frac{4}{1000}\times 300=0.8\times v_2{v}_2=1.5m/s$

now there is friction which resist the motion of block

friction retardation

$a=-\mu g=-0.3\times 10$

$a=-3m/s^2$

by using newtons law of motion

$v^2=u^2-2aS(v=0finalvelocityofblock)0={\left(1.5\right)}^2-2\times 3\times Ss=0.375m$

Option B is correct.