Question

A 1kg mass projected in air has speeds 3m/s and 4m/s at two different points of its trajectory.Of at these points ,the velocity vectors are normal to each other,then the least kinetic energy of the particle is

Answer 1

Given:

The mass of the body in air is  1 Kg.

The speed are 3 m/sec and 4 m/sec.

To find:

The least kinetic energy of the particle.

Solution:

The velocity in the projectiles are as

v=v'=u.cosФ

Now taking the resultant components.

3.cosФ=4.cos(90°-Ф)

3.cosФ=4.sinФ

tanФ=3/4

Applying trignometry formula.

sec²Ф-1=9/16

sec²Ф=25/16

or cosФ=4/5

Velocity on the x-component is =3×4/5=12/5

Now the kinetic energy

K.E=mv²/2

Putting the above value

K.E.=1×144×2/2×25

K.E=5.76 J

So the least kinetic energy of the particle is 5.76 J