Question

A bullet of mass $10kg$ strikes a sand bag at a speed of ${10^3}m{s^{ - 1}}$and gets embedded after travelling $5cm$ .Calculate
a. Resistance force exerted by the sand on the bullet.
b. Time taken by the bullet to come to rest.

Answer 1

Hint: We can solve this question using one-dimensional equations of motion. The main concept behind which we are going to use here is acceleration and velocity.
A force is defined as push or pull on an object due to the object's interaction with another object. Whenever there is an interaction between two objects, then we can say there is a force upon each of the objects.
Acceleration can be defined as the rate of change of velocity.
Velocity defines the direction of the movement of a particular body or object.
Here we are asked to calculate resistance force and time taken for the bullet to come to rest so for that first we need to find the acceleration. Then we can easily find the time taken and the resistance force.

Complete Step by step solution:
The formula which we are going to use here is
${v^2} = {u^2} + 2 \times a \times s$
Where v is the final velocity
U is the initial velocity
a is the acceleration
s is the distance covered.
The given values are
$m = 10kg$ $v = 0$ $u = {10^3}ms{}^{ - 1}$ $s = 5cm$
Now what I have to do is make the units the same. Here mass is in kilogram and distance is in metre. We convert it into gram and centimetre respectively.
$ \Rightarrow m = 10 \times {10^{ - 3}}g$
$ \Rightarrow m = {10^{ - 2}}g$
Similarly,
$s = 5 \times {10^{ - 2}}m$
Then substitute all the values in the above equation
$ \Rightarrow 0 = {\left( {{{10}^3}} \right)^2} + 2 \times a \times 5 \times {10^{ - 2}}$
$ \Rightarrow a = - {10^7}m{s^{ - 2}}$
We got the value of acceleration but we need to find time so substituting in the following equation will give time
$v = u + at$
$ \Rightarrow 0 = {10^3} - {10^7} \times t$
$ \Rightarrow t = {10^{ - 4}}s$.

b. The time taken by the bullet to come to rest is ${10^{ - 4}}s$
a. Resistive force $F = m \times a$
$ \Rightarrow F = {10^{ - 2}} \times {10^7}$
$ \Rightarrow F = {10^5}N$
$ \Rightarrow F = 100kN$

Additional information:
One dimensional motion describes the motion of objects which are moving in a straight line.
For example when a car is moving along a straight line.

Note:
There are four different types of motion. They are linear motion, rotary motion, reciprocating motion and oscillating motion.
ROTARY MOTION - Rotary motion is basically anything that moves in a circle.
OSCILLATING MOTION - Something that oscillates both back and forth.
LINEAR MOTION - Linear motion is anything that moves in a straight line.
Reciprocating motion is a repetitive up-and-down or back-and-forth linear motion.