Question

A bullet of mass \[20g\] and moving with $600m/s$ collides with a block of mass \[4kg\] hanging with the string. What is the velocity of the bullet when it comes out of the block, if the block rises to height $0.2m$ after collision.
A.$200m/s$
B.$150m/s$
C.$400m/s$
D.$300m/s$

Answer 1

Hint: Momentum, for any moving body, is just the product of its mass and its velocity. Traditionally, symbolized as ‘$p$‘, it could be formulated as$p=mv$and is also the first derivative of force with respect to time i.e. Force can also be equated as rate of change of momentum

Complete answer:
The concept of Conservation of Energy and momentum will be used here in the given question.
Initially the bullet has kinetic energy which gets divided into potential energy of the block and the kinetic energy of the bullet after collision.
Let the velocity of bullet and block after collision be ${{V}_{2}}\,\,and\,\,{{V}_{3}}$.
Initial momentum of the bullet =$mv$
Putting the values,
$\Rightarrow \dfrac{20}{100}\times \left( 600 \right)=12kgm/s$
Using conservation of momentum,
$12=\dfrac{20}{100}\times {{V}_{2}}+4{{V}_{3}}.........\left( 1 \right)$
Potential energy of block=$mgh$
$\Rightarrow 4\times 10\times 0.2=8J$
Using conservation of energy for block
Change in kinetic energy =work done
$\dfrac{1}{2}\times 4\times V_{3}^{2}=8$
${{V}_{3}}=2m/s$
Putting this value in $\left( 1 \right)$
\[{{V}_{2}}=\dfrac{4}{\left( \dfrac{20}{1000} \right)}=200m/s\]

Hence, the correct option would be (A) $200m/s$.

Note :
One is advised to remember the Law of conservation of momentum which is used in the above solution, which states that within some problem domain, the amount of momentum remains constant that is, just like mass momentum can neither be created nor destroyed, but can only be changed through the action of forces as described by Newton's laws of motion.