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 0.028 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 0.048 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of puck A. angle a=65 angle b=37

https://general.physics.rutgers.edu/gifs/CJ/7-13.gif

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.028 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 0.048 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of puck A. angle a=65 angle b=37

Explanation / Answer

A= 0.028 kg
B=0.048 kg

A*v(a)sin 65 = B*v(b) sin 37 (vertical direction)

Therefore:

v(a) = (B*v(b) sin 37) / A* sin 65) .... (eq1


A*v(a) cos 65 + B*v(b) cos 37 = 0.028*5.5 = 0.154 (eq2 (horizontal direction)

put(eq1 in (eq2

(B*v(b) sin 37) / sin 65) cos 65 + B*v(b) cos 37 = 0.154


(0.048*v(b) sin 37) / sin 65) cos 65 + 0.048*v(b) cos 37 = 0.154

0.01347* v(b) + 0.03833 *v(b) = 0.154

v(b) = 0.154/ (0.01347+0.03833)

v(b)=0.154/(0.0518)
v(b) = 2.972 m/s............................... Submit in eq1):

v(a) = (0.048*2.972 sin 37) / 0.028* sin 65)

v(a) = 3.3831 m/s

(a) Final speed of puck A is 3.3831 m/s

(b) Final speed of puck B is 2.972 m/s

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