Question

How much kinetic energy is lost in the collision? (See full question below)
A snooker ball having a mass of 0.04kg and initially moving with a speed of \[2m{\sec ^{ - 1}}\] strikes a stationary ball of the same mass. After the collision, the two balls are both moving at equal speeds at equal angles $30^\circ $ to the original direction of the incident ball. How much kinetic energy is lost in the collision?

Answer 1

Hint: The kinetic energy of the body is the energy possessed by the body by the virtue of its motion.
The law of energy states that energy can neither be created nor be destroyed but only can be converted from one form to another or can be transferred from one body to another body.
The Law of conservation of momentum states that momentum before the collision is equal to momentum after the collision.

Complete step by step answer:
Step 1:
Now in the question, it is given that this ball A strikes with a Stationary ball of the same mass and after the collision, both balls move with the same velocity.
Now here we will apply the law of conservation of momentum which states that momentum before the collision is always equal to momentum after Collision
We know that the momentum of the body is given by the product of its mass and velocity
Now, momentum before Collision will be equal to the sum of individual momentums of the two balls
${p_i} = {m_A}{u_A} + {m_B}{u_B}$
Where ${m_A}{u_A}$ is the initial momentum of ball A and ${m_B}{u_B}$ be the initial momentum of the other ball let’s say B
Now in the question, it is given that both balls have the same mass i.e, ${m_A} = {m_B} = m$
And ball B is stationary which means its initial velocity is zero i.e. \[{u_B} = 0\]
Substituting values, we get
${p_i} = \left( {0.04 \times 2} \right) + \left( {0.04 \times 0} \right)$
$ \to {p_i} = 0.08kgm{\sec ^{ - 1}}$
Step 2:
The final momentum will also be the sum of individual momentums of the body and will be calculated as
${p_f} = {m_A}{v_A} + {m_B}{v_B}$
${v_A},{v_B}$are final velocities after collision
Now as it is given in the question that both balls have the same mass i.e, ${m_A} = {m_B} = m$
And now both the balls have the same final velocities i.e, ${v_A} = {v_B} = v$
So, the final momentum will be
${p_f} = mv + mv$
$ \to {p_f} = 2\left( {0.04} \right)v$
$ \to {p_f} = 0.08v$
Step3:
Now we have got both initial as well as final momentum and according to the law of conservation of momentum we know that initial momentum and final momentum are equal so equating both we get
$\because {p_i} = {p_f}$
$ \to 0.08 = 0.08v$
$\therefore v = 1m{\sec ^{ - 1}}$
So, the final velocity of both the ball will be $1m{\sec ^{ - 1}}$
Step 4:
To calculate the change in kinetic energy we will subtract initial kinetic energy from final kinetic energy
Mathematically
$\Delta K.E = {\left( {K.E} \right)_f} - {\left( {K.E} \right)_i}$
$ \to \Delta K.E = \left( {\dfrac{1}{2}m{v^2}} \right) - \left( {\dfrac{1}{2}m{u^2}} \right)$
$ \to \Delta K.E = \left( {\dfrac{1}{2} \times 0.04 \times 1} \right) - \left( {\dfrac{1}{2} \times 0.04 \times 4} \right)$
$ \to \Delta K.E = \left( {\dfrac{1}{2} \times 0.04 \times 1} \right) - \left( {\dfrac{1}{2} \times 0.04 \times 4} \right)$
$ \to \Delta K.E = \left( {0.02} \right) - \left( {0.08} \right)$
$ \to \Delta K.E = - 0.06j$
Final answer:
The change in K.E is 0.06j and the kinetic energy is transferred to another ball as well hence it is reduced.

Note:
The negative sign shows a decrease in K.E because after the collision some energy is transferred to the other body to make it move
The total momentum before the collision is always equal to the momentum after the collision through individual momentums of the bodies may change.
The angle of $30^\circ $ is given here to confuse because it has got nothing to do with the answer.