Question

A $4g$ bullet is fired horizontally with a speed of $300m/s$ into $0.8$ kg block of wood 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. $0.19m$
• B. $0.375m$
• C. $0.569m$
• D. $0.758m$

Answer 1

Answer: (B)

$0.375m$

Given,

$m_1=4g=0.004kg$

$u=300m/s$

$m_2=0.8kg$

$\mu =0.3$

$v=?$

Applying momentum conservation on block and bullet

$m_{1}u=m_{2}v$

$v=\frac{m_{1}u}{m_2}=\frac{0.004\times 300}{0.8}$

$v=1.5m/s$

Retardation acceleration, $a=-\mu g=-0.3\times 10=-3m/s^2$

From 3rd equation of motion,

$2as=v^2-u^2$

$-2\times 3\times s=0^2-{1.5}^2$

$s=0.375m$

The correct option is B.