Question

A block of mass $10\text{kg}$ rests on a horizontal table. When it is hit by a bullet of mass $50\text{g}$ moving with speed $V$, it gets embedded in the block. The block moves , and comes to rest after moving a distance of $2\text{m}$ on the table. If a freely falling object were to acquire speed $\frac{V}{10}$ after being dropped from a height $H$, then neglecting energy losses and taking $g=10\text{m}/{\text{s}}^2$ find the value of $H$.
[Take coefficiant of friction between the block and the table to be $0.05$]

• A. $0.05\text{km}$
• B. $0.02\text{km}$
• C. $0.03\text{km}$
• D. $0.04\text{km}$

Answer 1

The correct option is D $0.04\text{km}$

The given situation is shown in figure.


Let $m_1=50\text{g}$ and $m_2=10\text{kg}$ and velocity of the bullet be $V$.

Using law of conservation of linear momentum before and after collision of bullet with block we get,

$m_1{v}_1+m_2{v}_2=(m_1+m_2)V_o$

$\Rightarrow 0.05\times V+10\times 0=(0.05+10)V_o$

$\Rightarrow V_o=\frac{V}{201}.......\left(i\right)$

Now kinetic friction between block and table will oppose the relative motion of the block on table.

$f=\mu m_{2}g$

So, retarding accelaration,$a=\frac{f}{m_2}=\mu g=0.05\times 10=0.5\text{m}/{\text{s}}^2$

Further, block stops after travelling $2\text{m}$ from its initial position.

$\therefore v^2-u^2=2aS$
$\Rightarrow 0^2-V_0^2=2as$
$\Rightarrow 0^2-{\left(\frac{V}{201}\right)}^2=-2\times 0.5\times 2$

$\Rightarrow V=284.3\text{m}/\text{s}$

For freely falling object, initial speed $u=0\text{m}/\text{s}$

Final speed , $v=\frac{V}{10}=28.43\text{m}/\text{s}$

Further, applying law of conservation of machanical energy we get,


$mgH+0=\frac{1}{2}m\times (28.43{)}^2+0$

$\Rightarrow H=\frac{(28.43{)}^2}{2\times 10}$

$=40.4\text{m}$

$=0.04\text{km}$

Hence, option (d) is the correct answer.

Why this question? Key Idea- This question is perfect blend of friction, free fall, momentum conservation and energy conservation.