Question

A projectile of mass 2 kg has velocities 3 mIs and 4 m/s at two points during its flight in the uniform gravitational field of the earth. If these two velocities are perpendicular to each other, then find the minimum kinetic energy of the particle during its flight.

Answer 1

Explanation:

Answer: The minimum kinetic energy of the particle during its flight is 25 J.

Explanation:

Given that,

Mass m = 2kgm=2kg

Velocity of first particle v_{1} = 3 m/sv

1

=3m/s

Velocity of second particle v_{2} = 4 m/sv

2

=4m/s

When the velocities of both particles are perpendicular to each other

Then, according to figure

Using Pythagoras's theorem

(CB)^{2} = (AC)^{2}+(AB)^{2}(CB)

2

=(AC)

2

+(AB)

2

(CB)^{2} = (3)^{2}+(4)^{2}(CB)

2

=(3)

2

+(4)

2

(CB)^{2} = 25(CB)

2

=25

CB = 5m/sCB=5m/s

The kinetic energy is

K.E= \dfrac{1}{2}mv^{2}K.E=

2

1

mv

2

Now, put the value of mass and velocity

K.E = \dfrac{1}{2}\times 2kg\times 25m/sK.E=

2

1

×2kg×25m/s

K.E = 25 JK.E=25J

Hence, the minimum kinetic energy of the particle during its flight is 25 J.