Question

A boy releases a $0.5kg$ ball on the frictionless floor with the speed of $20m{s^{ - 1}}$. The ball gets deflected by an obstacle on the way. After deflection it moves with $5$ percent of its initial kinetic energy. What is the speed of the ball now ?
(A) $14.41m{s^{ - 1}}$
(B) $1.00m{s^{ - 1}}$
(C) $19.00m{s^{ - 1}}$
(D) $4.47m{s^{ - 1}}$

Answer 1

Hint: In order to solve this question, we will first calculate the initial kinetic energy of the ball then we will equate it with the final kinetic energy which is five percent of the initial kinetic energy, and then we will solve for the final velocity of the ball.

Formula used:
If m is the mass of the body and v is the velocity of the body then Kinetic energy K.E is given by $K.E = \dfrac{1}{2}m{v^2}$

Complete answer:
According to the question, we have given that mass of the ball is $m = 0.5kg$ and while dropping the initial velocity of the ball was $u = 20m{s^{ - 1}}$ so initial kinetic energy of the ball was
$
  K.{E_i} = \dfrac{1}{2}(0.5){(20)^2} \\
  K.{E_i} = 100J \to (i) \\
 $

Now, after hitting the floor, the ball moves with $5$ percent of its initial kinetic energy so the final kinetic energy of the ball is given by
$
  K.{E_f} = \dfrac{5}{{100}} \times K.{E_i} \\
  K.{E_f} = \dfrac{5}{{100}} \times 100 \\
  K.{E_f} = 5J \\
 $

Now, the mass will remain the same which is $m = 0.5kg$ and let final velocity be $v$ so we get,
$
  K.{E_f} = \dfrac{1}{2}m{v^2} \\
  5 = \dfrac{1}{2}(0.5){v^2} \\
  v = \sqrt {20} \\
  v = 4.47m{s^{ - 1}} \\
 $


So, final velocity of the ball will be $4.47m{s^{ - 1}}$

Hence, the correct answer is option (D) $4.47m{s^{ - 1}}$

Note: It should be remembered that mass does not change only in the case of non-relativistic speeds and the SI unit of energy is called Joules denoted by J. and in mechanical collisions a body always lost some energy in the form of heat, sound, and other forms that’s why the kinetic energy of a body decreased after collisions but there is no loss in kinetic energy during elastic collisions.