Question

A bullet of mass $0.012\text{kg}$ and horizontal speed $70{\text{ms}}^{-1}$ strikes a $0.4\text{kg}$ block of wood kept at rest and suspended from the ceiling by means of thin wires. If on stricking the block, the bullet instantly comes to rest with respect to the block, then calculate the height to which the block rises. Also estimate the amount of heat produced in the block.

• A. $0.212\text{m},28.54\text{J}$
• B. $1.212\text{m},28.54\text{J}$
• C. $0.212\text{m},48.54\text{J}$
• D. None

Answer 1

The correct option is A $0.212\text{m},28.54\text{J}$

Here,
Mass of bullet $\left(m\right)=0.012\text{kg}$
Mass of block $\left(M\right)=0.4\text{kg}$
Speed of the bullet $\left(u\right)=70{\text{ms}}^{-1}$
Let $v$ be the velocity of the [block+bullet] system after collision
Now, using law of conservation of linear momentum, we can write that, $mu=(M+m)v$
or $v=\left(\frac{mu}{M+m}\right)$
$=\left(\frac{0.012\times 70}{0.4+0.012}\right)=\frac{0.84}{0.412}$
$=2.04{\text{ms}}^{-1}$
Thus, the speed of bullet + block system after collision is $2.04{\text{ms}}^{-1}$.
Let $h$ be the height measured from the ground up to which the block rises after striking.
Applying law of conservation of energy,
Increase in P.E. of the Bullet+Block system = Decrease in K.E. of the Bullet + Block system
Hence, we can write that,
$(M+m)gh=\frac{1}{2}(M+m)v^2$
or $h=\frac{v^2}{2g}=\frac{(2.04{)}^2}{2\times 9.8}$
$=0.212\text{m}$

Heat produced = Loss in K.E of Bullet + Block system.
$=\frac{1}{2}m{u}^2-\frac{1}{2}(M+m)v^2$
$=\frac{1}{2}\left(0.012\right)\times (70{)}^2-\frac{1}{2}\times 0.412\times (2.04{)}^2$
$=28.54\text{J}$
Thus, option (a) is the correct answer.