Question

Using second law of motion, derive the relation
between force and acceleration. A bullet of
10 g strikes a sand-bag at a speed of 103msland
gets embedded after travelling 5 cm. Calculate
(i) the resistive force exerted by the sand on
the bullet
(ii) the time taken by the bullet to come to rest.​

Answer 1

Answer:

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 p

1 =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 or

a= mF

v=u+at

0=103−107 t107

t=103

t=107103=10−4s

Answer 2

Answer:

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