Question

A projectile of mass 2kg has velocities 3m/s and 4m/s at two points during its flight if these two velocities are perpendicular to each other then find the minimum kinetic energy of the particle during its flight

Answer 1

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

Explanation:

Given that,

Mass $m = 2kg$

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

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

When the velocities of both particles are perpendicular to each other

Then, according to figure

Using Pythagoras's theorem

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

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

$(CB)^{2} = 25$

$CB = 5m/s$

The kinetic energy is

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

Now, put the value of mass and velocity

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

$K.E = 25 J$

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

Answer 2

"Given:

Mass = 2kg

Velocities = 3 m/sec & 4 m/sec; Velocities are perpendicular to each other

Minimum Kinetic Energy of the particle = ?

Solution:

As shown in the figure, this forms a right-angle triangle.

Thus, using Pythagoras's theorem,

$CB^2= AC^2 +AB^2$

$CB^2 = 32 + 42$

$CB^2 = 25$

CB = 5 m/sec;

And the Kinetic Energy is

$K.E = \frac {1}{2} mv^2$

On substituting,

$K.E = \frac {1}{2} (2 \times 25)$

= 25 J

Minimum Kinetic Energy of the particle is 25J."