Question

Using the IInd law of motion find the relation b/w acceleration and force.
Bullet=10g
Distance=5cm
initial velocity=10m/s
Final velocity=0m/s
So this bullet with velocity 10m/s gets embedded in sand after travelling 5cm into it
calculate the resistance force applied by the sand on the bullet
Time taken for the bullet to stop

Answer 1

The second law of motion states that rate of change of momentum of an object is proportional to the applied force...

Derivation:

Let initial and final momentum of the object are p1=mu and p2=mv respectively.

The change in momentum p=mv−mu

pnm×(v−u)

The rate of change of momentum is directly proportional to applied force F

i.e., F∞tm(v−u)

F=ktm(v−u)

F=k.m.a

The unit of force is so chosen that the value of the constant, k becomes one.

For this, one unit of force is defined as the amount that produced an acceleration of 1 ms−2

In an object of 1 kg mass. That is,

1 unit of force =k×(1 kg)×(1 m s)

Thus, the value of k becomes 1.

So we can write

F=ma

a=mF

v=u+at

• 0=103−107t

107t=103

t=107103

• =10−4s