Question

A bullet of mass 20g is fired with a velocity of 1200ms
-1
. After passing through a mud wall of 2m thick,
its velocity decreases to 200ms
-1
. What is the average resistance offered by the mud wall?

Answer 1

$\: mass \times retrdtion = \frac{20}{1000} \frac{ {1200}^{2} - {200}^{2} }{2 \times 2}$

this is average resistance offered by wall

Answer 2

Answer: Average resistance = 20kg m/s

Explanation:

mass of bullet = 20 grams

                       =20/1000 kilograms(converting to kg for the below formula)

                       =1/50 kilograms

initial velocity=1200m/s

momentum(kg m/s)=mass(kg) * velocity(m/s)

                                  =1/50 * 1200

                                  =24 kg m/s

therefore, initial momentum=24 kg m/s

velocity after hitting the wall=200m/s

mass=1/50 kg

momentum=mass * velocity

                   =1/50 * 200

                   =4 kg m/s

Finally, average resistance offered by the wall

=(initial momentum-final momentum) kg m/s

=(24-4)kg m/s

=20 kg m/s  (that's the answer!)

I hope my detailed explanation is clear to the readers