The drawing shows a collision between two pucks on an air-hockey table. Puck A h

Question

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 3.5 kg and is moving along the x axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 7.5 kg and is initially at rest. After the collision, the two pucks fly apart with the angles shown in the drawing.

5.+-6.75 points My Notes Ask Your Teacher (Note: Please be careful with roundoff error in this problem.) The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 3.5 kg and is moving along the x axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 7.5 kg and is initially at rest. After the collision, the two pucks fly apart with the angles shown in the drawing. 65° +5.5 mis 37 At rest Before collision After col lision (a) Find the final speed of: puck A: puck B: m/s m/s (b) Find the kinetic energy of the A+B system: before the collision: after the collision:

Explanation / Answer

a)

here by the conservation of momentum

ma * v0a = ma * vfa * cos(65deg) + mb * vfb * cos(37deg)

while momentum conservation in the y direction is

0 = ma * vfa * sin(65) - mb * vfb * sin37

vfb = ma * vfa * sin65 / (mb * sin37 )

put the vfb in first equation

ma * v0a = ma * vfa * cos65 + ( ma * vfa * sin65 / sin37 ) * cos37

vfa = v0a / (cos65 + (sin65 / tan37)

vfa = 5.5 / ( cos65 + (sin65 / sin37) = 3.4 m/s

then

vfb = 3.5 * 3.4 * sin65 / ( 7.5 * sin37) = 2.389 m/s

b)

0.5 * ma * v0a^2 = 0.5 * ma * vfa^2 + 0.5 * mb * vfb^2

put the values in the formula

before the collision = after the collision

(0.5 * 3.5 * 5.5^2) = (0.5 *3.5 * 3.4^2 + 0.5 * 7.5 * 2.389^2)

before collision = 52.93 J

after collision = 41.63 J

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