Question

Consider the collision of two identical hockey pucks on a smooth ice rink. Various but not all components of the velocities of the two pucks before and after the collisions have been determined and are recorded in the table below.

Consider the collision of two identical hockey pucks on a smooth ice rink. Various but not all components of the velocities of the two pucks before and after the collisions have been determined and are recorded in the table below. vy (in cm/s) after collsion c) What percentage of kinetic energy is lost in the collision? vx (in cm/s) after collsion b) Puck 2,

Assume that friction with ice is negligible. Complete the missing entries.

a) Puck 2, $v_{x}$ (in cm/s) after collsion

b) Puck 2, $v_{y}$ (in cm/s) after collsion

c) What percentage of kinetic energy is lost in the collision?

Answer 1

Given that

the two pucks are identical

Puck	Before $v_x$	Before $v_y$	After $v_x$	After $v_y$
1	21	35	-13	10
2	-46	15

Conserving the linear momentum before and after collision
let the mass of each puck be 'm' then

Before collision :
Along X axis :momentum   $P_{x_1}=21m-46m=-25m$

Along Y axis :momentum   $P_{y_1}=35m-15m=50m$

After collision :

Along X axis :momentum $P_{x_2}=v_xm-13m$

Along Y axis :momentum $P_{y_2}=v_ym+10m$

But ,
   $P_{x_2}=P_{x_1}\Rightarrow v_xm-13m=-25m$

   $v_x=-12\: cm/s$

similarly
     $P_{y_2}=P_{y_1}\Rightarrow v_ym+10m=50m$

$v_y=40\: cm/s$

$K.E_1=\frac{m(21^2+35^2)+m(46^2+15^2)}{2}=\frac{4007m}{2}$

$K.E_2=\frac{m(13^2+10^2)+m(12^2+40^2)}{2}=\frac{2013m}{2}$

percentage of kinetic energy is lost in the collision,
     $\frac{K.E_1-K.E_2}{K.E_1}*100=49.76 \: percent$