Question

On a frictionless surface, a block of mass M moving at speed υ collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle $\theta $ to its initial direction and has a speed $\dfrac{v}{3}$. The second block's speed after the collision is:
(A) $\dfrac{\sqrt{3}}{2}\nu $
(B) $\dfrac{2\sqrt{2}}{3}\nu $
(C) $\dfrac{3}{4}\nu $
(D) $\dfrac{3}{\sqrt{2}}\nu $

Answer 1

Hint- When two bodies collide with each other, an elastic collision occurs, but in the total kinetic energy, there is no loss. In other words, it is an interaction between two bodies where the two bodies' overall kinetic energy is the same.

Complete step by step answer:
 Energy conservation, the physical theory according to which the energy of interacting bodies or particles remains unchanged in the closed system. Kinetic energy, or energy of motion, was the first sort of energy to be known. In such particles, termed elastic collisions, the sum of the particles' kinetic energy before the collision is equal to the sum of the particles' kinetic energy after the collision.
In the question it is given that the ball collides elastically and as we know that an elastic collision is a collision that retains all the kinetic energy, KE, and momentum, p.
 Since, $KE=\dfrac{1}{2}m{{v}^{2}}$
In the question it is clearly given that the moving ball comes to rest that implies Initial kinetic energy is equal to the final kinetic energy.
Therefore, by using law of conservation of kinetic energy we can say that,
\[\dfrac{1}{2}{m_1}v_1^2 + \dfrac{1}{2}{m_2}v_2^2 = \dfrac{1}{2}{m_1}v_1^{'2} + \dfrac{1}{2}{m_1}v_2^{'2}\]
From the above equation we can say that energy can neither be produced nor destroyed-it can only be transferred from one energy source to another. This means that, whether it's applied from the outside, a machine still has the same amount of electricity. In the case of non-conservative forces, in which energy is converted from mechanical energy into thermal energy, this is especially complicated, but the total energy stays the same. Transforming energy from one shape to another is the only way to use energy.
In the above equation we have to find the value of $v_{2}^{'2}$
Eliminate, $v {1}^{'2}$ as $\dfrac{v}{3}$
The block comes at rest finally so, $v_{2}^{2}=0$
Therefore we have,
$\dfrac{1}{2}{{m}_{1}}{{v}_{1}}^{2}+0=\dfrac{1}{2}{{m}_{2}}{{\left( \dfrac{v}{3} \right)}^{2}}+\dfrac{1}{2}{{m}_{2}}v_{2}^{2}$
Mass is the same so we can cancel out m from the above equation taking it as common.
$\Rightarrow {{v}^{2}}=\dfrac{{{v}^{2}}}{9}+v_{2}^{2}$
$\Rightarrow v_{2}^{2}=\dfrac{8{{v}^{2}}}{9}$
$\Rightarrow v_{2}^{{}}=\dfrac{2\sqrt{2}{{v}^{{}}}}{3}$

Therefore, the correct answer is option B.

Note: As it moves upward toward the force of gravity, the kinetic energy expended by a body slowing down was known to be transformed into potential energy, or stored energy, which in essence is transformed back into kinetic energy as the body speeds up on its return to Earth. For one, kinetic energy is converted to potential energy when a pendulum swings upward.