The drawing shows a collision between two pucks on an air-hockey table. Puck A h
ID: 2258501 • Letter: T
Question
The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.0450 kg and is moving along the x axis with a velocity of +5.71 m/s. It makes a collision with puck B, which has a mass of 0.0900 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 speed of (a) puck A and (b) puck B.
Explanation / Answer
A= 0.0450 kg
B=0.0900kg
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 = 0.0450*5.71 = 0.257 (eq2 (horizontal direction)
put(eq1 in (eq2
(B*v(b) sin 37) / sin 65) cos 65 + B*v(b) cos 37 = 0.257
(0.09*v(b) sin 37) / sin 65) cos 65 + 0.09*v(b) cos 37 = 0.257
0.025256* v(b) + 0.071877 *v(b) = 0.257
v(b) = 0.257/ (0.025256+0.071877
v(b)=0.257/(0.097133)
v(b) = 2.6458 m/s............................... Submit in eq1):
v(a) = (0.09*2.6458 sin 37) / 0.045* sin 65)
v(a) = 3.5137 m/s
(a) Final speed of puck A is 3.5137 m/s or 3.51m/s(approx)
(b) Final speed of puck B is 2.6458 m/s or 2.64m/s(approx)
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