Question

In a pool game, a 1 kg pool ball moving at \[ + 10\,m/s\] strikes a 2 kg pool ball that is initially at rest. Identify which statement is true immediately after the elastic collision?
The 1 kg pool ball be at rest.
(A) The 2 kg pool ball must have a negative velocity.
(B) The 2 kg ball’s speed must be less than 10 m/s.
(C) The two pool ball sticks together.
(D) The total kinetic energy of the two pool balls must be equal to 100 J.

Answer 1

Hint: From the equation of conservation of momentum and conservation of kinetic energy, express the velocity of the second pool ball of mass 2 kg. Determine its velocity and observe whether it is less than 10 m/s. We know that the coefficient of restitution is equal to 1 in the elastic collision.

Complete step by step answer:
We know that in elastic collision, the linear momentum of the two colliding bodies conserves. Also, the kinetic energy of the bodies also conserves. The conservation of momentum implies that the initial momentum and final momentum of the system of two colliding bodies remains the same.
Therefore, we can write,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\] …… (1)
Here, \[{m_1}\] is the first colliding body, \[{m_2}\] is the mass of second colliding body, \[{u_1}\] is the initial velocity of the first body, \[{u_2}\] is the initial velocity of the second body, \[{v_1}\] is the final velocity of the first body and \[{v_2}\] is the final velocity of the second body.
Also, from the conservation of kinetic energy we can write,
\[\dfrac{1}{2}{m_1}u_1^2 + \dfrac{1}{2}{m_2}u_2^2 = \dfrac{1}{2}{m_1}v_1^2 + \dfrac{1}{2}{m_2}v_2^2\] …… (2)
From equation (1) and (2), after further simplification, we can express the velocity of the second particle as follows,
\[{v_2} = \dfrac{{2{m_1}{u_1} + {u_2}\left( {{m_2} - {m_1}} \right)}}{{{m_1} + {m_2}}}\]
We have given that the second pool ball is initially at rest. Therefore, we can substitute\[{u_2} = 0\] in the above equation.
\[{v_2} = \dfrac{{2{m_1}{u_1}}}{{{m_1} + {m_2}}}\]
We substitute 1 kg for \[{m_1}\], 2 kg for \[{m_2}\] and 10 m/s for \[{u_1}\] in the above equation.
\[{v_2} = \dfrac{{2\left( 1 \right)\left( {10} \right)}}{{1 + 2}}\]
\[ \Rightarrow {v_2} = \dfrac{{20}}{3}\]
\[ \Rightarrow {v_2} = 6.67\,m/s\]
Therefore, we can see that velocity of the second pool ball of mass 2 kg is less than 10 m/s.
So, the correct answer is option (C).
We have to verify other options as well to answer this question.
We know that, In an elastic collision, the colliding body does not come to the rest if its mass is less than the other body. Therefore, the option (A) is incorrect.
We have given that after the collision, both bodies move in the same direction in an elastic collision. So, the velocity of a 2 kg pool ball will not be negative. Therefore, the option (B) is incorrect.We know that the two colliding bodies stick together in perfectly inelastic collisions and not in elastic collisions. Therefore, the option (D) is incorrect.
Let’s calculate the kinetic energy of the two balls before the collision as the kinetic energy after the collision is the same as kinetic energy before the collision.
\[{\left( {K.E} \right)_i} = \dfrac{1}{2}{m_1}u_1^2 + \dfrac{1}{2}{m_2}u_2^2 = \dfrac{1}{2}\left( 1 \right){\left( {10} \right)^2} + \dfrac{1}{2}\left( 2 \right){\left( 0 \right)^2}\]
\[ \Rightarrow {\left( {K.E} \right)_i} = \dfrac{1}{2}\left( 1 \right){\left( {10} \right)^2}\]
\[ \therefore {\left( {K.E} \right)_i} = 50\,J\]

Therefore, the kinetic energy is not equal to 100 J. Thus, the option (E) is incorrect.So, the only correct answer is option (C).

Note: It takes a lot of calculation to determine the velocity of the second body in the elastic collision. Therefore, students should remember the formula.Only in perfectly elastic collision, the total kinetic energy of the system before the collision remains the same after the collision. There is no loss and the coefficient of restitution is equal to one. But in reality, there is always some loss in the kinetic energy which gets converted into an internal energy. In this case, even if the collision is an elastic collision, the coefficient of restitution becomes slightly less than 1.