Question

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Answer 1

Answer:

In the physical sciences, a particle is a small localized object to which can be ascribed several physical or chemical properties such as volume, density or mass.

Answer 2

Topic :-

Kinetic Energy

Given :-

A particle of mass 'm' has half the Kinetic Energy of another particle of mass 'm/2'.If the speed of heavier particle is increased by 2m/s, its new Kinetic Energy equals the original Kinetic Energy of the lighter particle.

To Find :-

The original speed of heavier particle.

Formula to be Used :-

$Kinetic\:Energy=\dfrac{1}{2}mv^2$

where,

• m = Mass of Particle

• v = Velocity of Particle

Solution :-

Let us assume that velocity of heavier particle ( mass 'm' ) is 'v'.

So, its Kinetic Energy is,

$K.E._1=\dfrac{1}{2}mv^2$

Let us assume that velocity of lighter particle ( mass 'm/2' ) is v'.

So, its Kinetic Energy is,

$K.E._2=\dfrac{1}{2}\dfrac{m}{2}(v')^2$

$K.E._2=\dfrac{1}{4}m(v')^2$

Now,

It is given that,

$K.E._1=\dfrac{1}{2}K.E._2$

$\dfrac{1}{2}mv^2=\dfrac{1}{2}\left ( \dfrac{1}{4}m(v')^2 \right )$

$\dfrac{1}{2}mv^2=\dfrac{1}{8}m(v')^2$

This can be reduced to,

$v^2=\dfrac{v'^2}{4}$

$v=\dfrac{v'}{2}$

So, velocity of heavier particle is half of velocity of lighter particle.

Now,

Speed of heavier particle is increased by 2m/s that is :

( v + 2 ) m/s

New Kinetic Energy of heavier particle,

$K.E._3=\dfrac{1}{2}m(v+2)^2$

It is given that it is equal to $K.E._2.$

$K.E._3=K.E._2$

$\dfrac{1}{2}m(v+2)^2=\dfrac{1}{4}m(v')^2$

Cross multiply after cancelling 'm'.

$2(v+2)^2=(v')^2$

$2v^2+8v+8=(v')^2$

Now, replace v' with 2v.

$2v^2+8v+8=(v')^2$

$2v^2+8v+8=(2v)^2$

$2v^2+8v+8=4v^2$

$2v^2-8v-8=0$

$v^2-4v-4=0$

Solve the quadratic equation using Quadratic Formula,

$v=\dfrac{4 \pm \sqrt{16-4(-4)}}{2}$

$v=\dfrac{4 \pm 4\sqrt{2}}{2}$

$v=2 \pm 2\sqrt{2}$

Speed can't be negative, so we reject negative value of 'v'.

Answer :-

So, the speed of heavier particle is,

$v=2(1+ \sqrt{2})$ m/s

which is option B.