Question

Two identical pucks collide on an air hockey table. One puck was originally at rest.

(a) If the incoming puck has a speed of\(6.00 m/s \)andscatters to an angle of ,what is the velocity (magnitude anddirection) of the second puck? (You may use the result that \({\theta _1}-{\theta _2} = 90°\) for elastic collisions of objects that have identicalmasses.)

(b) Confirm that the collision is elastic.

Short answer

(a)The velocity of the pucks after Collision is\(5.2 m/s\)and\(3 m/s\).

(b) Because energy is conserved so the collision is elastic.

Step-by-step solution

Definition of collision

According to collision theory, only a specific number of collisions between acceptable reactant particles with the correct orientation result in a detectable or noticeable change; these successful modifications are referred to as successful collisions.

Given data

(a) The puck data before the collision is as shown in the below figure:

The above collision is elastic hence

\({P_{1b}} + {P_{2b}} = {P_{1a}} + {P_{2a}}\)

The puck after the collision

\(\begin{aligned}{\theta _1} - {\theta _2} &= 90\\30 - {\theta _2} &= 90\\{\theta _2} &= - 60\end{aligned}\)

Hence figure for that will be as above .The moment components will be

Determine the X-component and Y-component

By putting all the value into the equation we get

X- components

\(\begin{aligned}{P_{1bx}} + {P_{2bx}} &= {P_{1ax}} + {P_{2ax}}\\{P_{1b}} + 0 &= {P_{1a}}cos30 + {P_{2a}}cos60\end{aligned}\)…………………(1)

Y- components

\(\begin{aligned}{P_{1by}} + {P_{2by}} &= {P_{1ay}} + {P_{2ay}}\\0 + 0 &= {P_{1a}}sin30 - {P_{2a}}sin60\\{P_{1a}} &= \dfrac{{{P_{2a}}sin60}}{{sin30}}\end{aligned}\)

Putting the above value in equation 1

\(\begin{aligned}{P_{1bx}} + {P_{2bx}} &= {P_{1ax}} + {P_{2ax}}\\{P_{1b}} &= \left( {\dfrac{{{P_{2a}}sin60}}{{sin30}}} \right)cos30 + {P_{2a}}cos60\\{P_{1b}} &= {P_{2a}}\left( {\left( {\dfrac{{sin60}}{{sin30}}} \right)cos30 + cos60} \right)\end{aligned}\)

\(\begin{aligned}\dfrac{{\cancel{m}{V_{1b}}}}{{\left( {\left( {\dfrac{{sin60}}{{sin30}}} \right)cos30 + cos60} \right)}} &= \cancel{m}{V_{2a}}\\\dfrac{{{V_{1b}}}}{{\left( {\left( {\dfrac{{sin60}}{{sin30}}} \right)cos30 + cos60} \right)}} &= {V_{2a}}\\\dfrac{6}{2} &= {V_{2a}}\\{V_{2a}} &= 3\,m/s\end{aligned}\)

The velocity of puck after the Collison is \(3 m/s\)

Determine the velocity of first puck

Similarly the velocity of the first puck is

\(\begin{aligned}{P_{1a}} &= \dfrac{{{P_{2a}}sin60}}{{sin30}}\\m{V_{1a}} &= \dfrac{{m{V_{2a}}sin60}}{{sin30}}\\{V_{1a}} &= \dfrac{{{V_{2a}}sin60}}{{sin30}}\\{V_{1a}} &= \dfrac{{\left( 3 \right)sin60}}{{sin30}} &= 5.2\,m/s\end{aligned}\)

The velocity of the pucks after Collision is \(5.2 m/s\)and\(3 m/s\).

Determine conservation of energy

b) Here if the conservation of energy is conserved than it is elastic collision.

\(K{E_i} = K{E_f}\)

\(\begin{aligned}\dfrac{1}{2}m{V_{1b}}^2 &= \dfrac{1}{2}m{V_{1a}}^2 + \dfrac{1}{2}m{V_{2b}}^2\\{V_{1b}}^2 &= {V_{1a}}^2 + {V_{2b}}^2\\{6^2} &= {\left( {5.2} \right)^2} + {3^2}\\36 &= 36\end{aligned}\)

Hence the kinetic energy is conserved.