Question

A bullet of mass $5\,g$ travels with a speed of $500\,m{{s}^{-1}}$.If it penetrates a fixed target which offers a constant resistive force of $1000\,N$ to the motion of the bullet, find:
(a) The initial kinetic energy of the bullet
(b) The distance through which the bullet has penetrated.
A. $s=0.625\,m$
B. $s=0.725\,m$
C. $s=0.225\,m$
D. $s=0.65\,m$

Answer 1

Here we will know what energy means and then in order to find the kinetic energy of the bullet we can find it directly by putting the values in the equation for kinetic energy, we can also find the distance penetrated by the bullet by using the formula for work done.

Complete step by step answer:
The ability of an object to perform work is referred to as its energy. When the object is in motion the energy possessed by the object is known as kinetic energy. When an object undergoes a change in its position (displacement) by an external force. The energy required to change that object’s position is called work done. Equations of motion are physics equations that define a physical system's action in terms of its motion as a function of time. There are in total three equations of motion.

(a) Here the given data are:
Mass= $$5,g$$$=\dfrac{5}{1000},kg$Speed=$500,m{{s}^{-1}}$Force=$1000,N$We know that kinetic energy is given as:$K.E=\dfrac{1}{2}m{{v}^{2}}$$v=500m/s$$\Rightarrow K.E=\dfrac{1}{2}\times \dfrac{5}{1000}\times 500\times 500$$\therefore K.E=625,J$

(b) The distance penetrated through bullet:
It is well known by us that work done is equal to energy here,
$\text{Work done=Force} \times \text{Displacement}$
$W=F\cdot s=K.E$
We have the resistive force equal to $1000N$
$1000\cdot s=625\,J$
$\therefore s=0.625\,m$

Hence, the correct answer is option A.

Note: Another method to get the distance through which bullet penetrated is:
We know that $\text{Force=mass} \times \text{acceleration}$
$1000=\dfrac{5}{1000}\times \text{acceleration}$
$\text{Acceleration}=1000\times \dfrac{1000}{5}=200000\,m/{{s}^{2}}$
Here we will use third equation of motion:
${{v}^{2}}={{u}^{2}}+2as$
${{\left( 500 \right)}^{2}}=0+2\times 200000\times s$
$250000=400000\times s$
$\therefore s=0.625\,m$
We did not approach this method as it is lengthier than the one we did above. Indeed one can solve the problem by their preferable choice of method but one should approach the less time consuming method.