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A 0.050 kg air puck moving with a velocity of 2.0 m/s collides with an identical but stationary air puck. The direction of the incident puck is changed by 60o from its original path and the angle between the two pucks after the collision is 90 o. What are the speeds of the two pucks after they collide?
- oldschoolLv 72 months agoFavorite Answer
The initial momentum Pio = 0.1 = the final momentum, vector sum Pi + Ps
Pi*cos60 +Ps*cos30 = 0.1 = Pi/2 + √3 *Ps/2 AND 0.2 = Pi+ √3 *Ps = 0.2
The vertical component of the the initial momentum = 0 so the vertical components of the final momentum must cancel.
Ps*sin(30) + Pi*sin(-60) = 0 = Ps/2 - Pi*√3/2 = 0 SO Ps = Pi *√3
Substitute into Pi+ √3 *Ps = 0.2
Pi+ √3 *(Pi*√3) = 4*Pi = 0.2
Pi = 0.2/4 = 0.05 = 1/20 or 5 cm/s <<<
Ps = Pi *√3 = 0.0866 = √3/20 or about 8.66 cm/s <<<<
Pi*cos60 +Ps*cos30 = 0.05*cos60 +0.0866*cos30 = 0.0999978 close enough
Ps*sin(30) + Pi*sin(-60) = 0.0866*sin(30) + 0.05*sin(-60) ≈ 0 checks