Question

What percentage of K.E of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 4times its mass?

Answer 1

Hint: In order to answer this question, first we will write the velocities of both the given bodies after the collision, and then we will write the equation in terms of mass and velocities when the momentum is conserved, to find the change in kinetic energy of the body whose mass is 4times the mass of first body.

Complete step by step answer:
As per the question, a body has a mass $m$ and it's moving with a velocity $v$. And another body, whose mass is $4m$ and it is in a rest position. After collision, the body of mass $m$ is moving with a velocity of ${v_1}$ and the other body of mass $4m$ is moving with a velocity of ${v_2}$ .
So, the Kinetic Energy will be:
$E = \dfrac{1}{2}m{v^2}$
As we know, if the bodies collide, then the momentum is conserved.
$\because mv = m{v_1} + 4m{v_2}$
$ \Rightarrow v = {v_1} + 4{v_2}$ ……….eq(i)

Now, in the question, there isn’t mentioned the type of collision, so by default we will take the collision as an elastic collision.As after collision, the change in velocity occurs.
$e = 1 = {v_2} - {v_1}$
\[ \Rightarrow {v_2} - {v_1} = v\] ……….eq(ii)
Now, by adding eq(i) and eq(ii):-
$2v = 5{v_2} \\
\Rightarrow {v_2} = \dfrac{2}{5}v \\ $
Since, after the collision, ${E_2}$ or the K.E of another body of mass $4m$ will be:-
${E_2} = \dfrac{1}{2}(4m)(\dfrac{4}{{25}}){v^2} \\
\Rightarrow {E_2} = \dfrac{{16}}{{25}}(\dfrac{1}{2}m{v^2}) \\
\Rightarrow {E_2} = \dfrac{{16}}{{25}}E $
So, the energy transferred is $\dfrac{{16}}{{25}}E$. Therefore,
Kinetic Energy percentage $ = \dfrac{{\dfrac{{16}}{{25}}E}}{E} \times 100 = 64\% $

Hence, \[64\% \] of K.E of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 4 times its mass.

Note: Momentum, the product of the mass of a particle and its velocity. The total momentum of the two objects before and after a collision between object 1 and object 2 in an isolated system is equal to the total momentum of the two objects after the collision.