Sphere A is moving at a speed of 16 ft/s when it strikes sphere B which is at re
ID: 2994344 • Letter: S
Question
Sphere A is moving at a speed of 16 ft/s when it strikes sphere B which is at rest. The mass of each sphere is m and the horizontal surface is frictionless. The impact causes sphere B to break into two equal pieces each of mass m/2. 0.7 s after the collision one piece reaches point C and 0.9 s after the collision the other piece reaches point D. Determine: (a) the velocity of A after the collision, (b) the angle theta and the speed of the two pieces after the collision.
Sphere A is moving at a speed of 16 ft/s when it strikes sphere B which is at rest. The mass of each sphere is m and the horizontal surface is frictionless. The impact causes sphere B to break into two equal pieces each of mass m/2. 0.7 s after the collision one piece reaches point C and 0.9 s after the collision the other piece reaches point D. Determine: (a) the velocity of A after the collision, (b) the angle theta and the speed of the two pieces after the collision.Explanation / Answer
Two Pieces :
Horizontal velocity = distance/time
(v1x) = 6.3/0.7 = 9 ft/s
(v2x) = 6.3/0.9 = 7ft/s
Momentum concservation in x :
m*16 + m*0 = m/2*9 + m/2*7 + m*v
where v = speed of sphere A in x-direction
m*16 = m*8 + m*v
v = 8 ft/s
Now motion in y-direction :
Two pieces :
6.3* tan 30 = (v1y)*t- 0.5*9.8*t^2
t = 0.7 s
vy = 8.626 ft
Second piece
6.3* tan theta = (v2y)*0.9 - 4.9*0.9^2
v2y - 7 tan theta = 4.41 -------(1)
tan theta = 9* tan 30 / 7
theta = 36.586 degrees
v2y = 4.41 + 7 tan theta = 9.606
Momentum concservation in y :
m*0 + m*0 = m/2*8.626 + m/2*(v2y) + m*vy
where vy = speed of sphere A in y-direction
-8.626/2 - 9.606/2 = vy
vy = -9.116 m/s
1) speed of A = sqrt (vx^2+ vy^2) = sqrt( 8^2 + 9.116^2) = 12.13 ft/s
2) theta = 36.58 degrees
V(1) = 12.466 ft/s
V(2) = 11.886 ft/s
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