A body of mass $m = 0.10kg$ has an initial velocity of $3\hat {i}~m{s^{ - 1}}$. It collided elastically with another body, B of the same mass which has an initial velocity of $5\hat{j}~m{s^{ - 1}}$. After collision, A moves with a velocity $v = 4(\hat i + \hat j)~m{s^{ - 1}}$ The energy of B after collision is written as $\dfrac{x}{{10}}J$. The value of $x$ is

Hint: In order to solve this question, we will first calculate the final velocity of body B using the law of conservation of linear momentum and then we will use the Kinetic energy formula to calculate the kinetic energy of body B and hence calculate the value of x.

Formula Used:
According to the law of conservation of momentum:
Momentum before Collision=Momentum after collision
$\dfrac {{m_1}{{u_1}^2}}{2}+\dfrac {{m_2}{{u_2}^2}}{2}=\dfrac {{m_1}{{v_1}^2}}{2}+\dfrac {{m_2}{{v_2}^2}}{2}$
Where,
${u_1},~{u_2}$-The initial velocities of the body
${v_1},~{v_2}$-The initial velocities of the body

Complete answer:
We have given that,
Two bodies A and B have same mass ${m_A} = {m_B} = 0.10kg$
The initial velocity of body A=${u_A} = 3\hat {i}~m{s^{ - 1}}$
The initial velocity of body B=${u_B} = 5\hat jm{s^{ - 1}}$
So, the Initial Momentum of the system is
${P_i} = {m_A}{u_A} + {m_B}{u_B}$

on substituting the values we get,
${P_i} = 0.10(3\hat i + 5\hat j)~kgm{s^{ - 1}} \to (i)$

Now, after the elastic collision,
The a velocity of body A =${v_A} = 4(\hat i + \hat j)m{s^{ - 1}}$
Let the final velocity of body B is ${\vec v_B}$

then final momentum is given by
${P_f} = {m_A}{v_A} + {m_B}{v_B}$

on putting the values we get,
${P_f} = 0.10(4\hat i + 4\hat j + {\vec v_B})kgm{s^{ - 1}} \to (ii)$

According to the law of conservation of momentum, equate both equations (i) and (ii) we get,
$
  {P_i} = {P_f} \\
  0.10(3\hat i + 5\hat j) = 0.10(4\hat i + 4\hat j + {{\vec v}_B}) \\
  {{\vec v}_B} = ( - \hat i + \hat j)m{s^{ - 1}} \\
 $

So, velocity of body B after collision is ${v_B} = ( - \hat i + \hat j)m{s^{ - 1}}$ and its mass is ${m_B} = 0.10kg$ then, Kinetic energy is given by $K.E = \dfrac{1}{2}{m_B}{v_B}^2$ here, magnitude of velocity of body B is
$
  \left| {{v_B}} \right| = \left| {( - \hat i + \hat j)} \right|m{s^{ - 1}} \\
  \left| {{v_B}} \right| = \sqrt 2 m{s^{ - 1}} \\
 $

On putting the values we get,
$
  K.E = \dfrac{1}{2}(0.10){(\sqrt 2 )^2} \\
  K.E = 0.10J \\
 $

And we have given that K.E is written as $\dfrac{x}{{10}}J$ comparing this with the derived value $K.E = 0.10J$ we get,
$
  \dfrac{x}{{10}} = 0.10 \\
  x = 1 \\
 $

Hence, the value of x is 1.

Note: We should always remember that during an elastic collision there is no loss in kinetic energy. i.e., there is no loss in deformation potential energy. At the same time, It should be noted that given velocities of the body are in vector form so while solving such questions always add and subtract the components of vectors carefully.