Question

A metal ball of mass $2\,Kg$ moving with a velocity of $36\,Kmh{r^{ - 1}}$ has a head on collision with a stationary ball of mass $3\,Kg$ If after the collision, the two balls move together, the loss in Kinetic energy due to collision is:
A. $140\,J$
B. $100\,J$
C. $60\,J$
D. $40\,J$

Answer 1

Hint: In order to solve this question, we will first find the final velocity of the system when two balls stick together after the collision using principle of linear conservation of momentum and then we will calculate initial and final kinetic energy of the system and later will find the difference in Kinetic energy.

Formula Used:
Principle of linear conservation of momentum is written as
Initial momentum $ = $ Final momentum
${P_i} = {P_f}$
${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$
where, ${m_1},{m_2}$ denote the mass of two bodies, ${u_1},{u_2}$ denote the initial velocities of two bodies and ${v_1},{v_2}$ denote the final velocities of two bodies.
Kinetic energy as,
$K.E = \dfrac{1}{2}m{v^2}$
where $m$ is mass and $v$ is velocity.

Complete step by step answer:
According to the question, we have given that Before collision initial condition of system as
${m_1} = 2Kg$ and ${u_1} = 36\,Kmh{r^{ - 1}} = 10\,m{s^{ - 1}}$
$\Rightarrow {m_2} = 3Kg$ ${u_2} = 0$ So,
Initial momentum is
${P_i} = (2 \times 10) + (3 \times 0)$
where $P$ denotes momentum.
${P_i} = 20\,Kgm{s^{ - 1}}$
After collision both bodies move together, let V be the final velocity of system and mass as $m = 2 + 3 = 5Kg$
final momentum is ${P_f} = 5V$
From principle of conservation of momentum we have,
${P_i} = {P_f}$
Put the values we get,
$\Rightarrow 20 = 5V$
$\Rightarrow V = 4\,m{s^{ - 1}}$

Now, Initial Kinetic energy as
$K.{E_i} = \dfrac{1}{2}{m_1}{u_1}^2 + \dfrac{1}{2}{m_2}{u_2}^2$
Putting the values we get,
$\Rightarrow K.{E_i} = \dfrac{1}{2}2(100) + \dfrac{1}{2}3(0)$
$\Rightarrow K.{E_i} = 100J$
For, final kinetic energy as
$K.{E_f} = \dfrac{1}{2}m{V^2}$
Putting value we get,
$\Rightarrow K.{E_f} = \dfrac{1}{2}5(16)$
$\Rightarrow K.{E_f} = 40J$
So, Loss in kinetic energy is $K.{E_i} - K.{E_f}$ we get,
$\Delta K.E = 100 - 40$
$\therefore \Delta K.E = 60J$

Hence, the correct option is C.

Note: It should be remembered that the collision is not an elastic collision, because in elastic collision initial kinetic energy is equal to final kinetic energy here, energy is lost during collision so this type of collision may be considered as inelastic collision and the conversion unit is used as $1\,Kmh{r^{ - 1}} = \dfrac{5}{{18}}\,m{s^{ - 1}}$.