*7-7 Collisions in Two Dimensions *44. (II) Billiard ball A of mass mA = 0.120 k
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Question
*7-7 Collisions in Two Dimensions *44. (II) Billiard ball A of mass mA = 0.120 kg moving with speed VA = 2.80 m/s strikes ball B, initially at rest, of mass mB = 0.140 kg. As a result of the collision, ball A is deflected off at an angle of 30.0 degree with a speed vA' = 2.10 m/s. (a) Taking the x axis to be the original direction of motion of ball A, write down the equations expressing the conservation of momentum for the components in the x and y directions separately. (b) Solve these equations for the speed, vB', and angle, theta B', of ball B after the collision. Do not assume the collision is elastic.Explanation / Answer
In the x direction....
mAvA + mBvB = mAvA'(cos 30) + mBvB'(cos @B')
since vB is zero initially, that cancels...
mAvA = mAvA'(cos 30) + mBvB'(cos @B')
In the y direction, neithe mass has any initial momentum, so...
0 = mAvA'(sin 30) - mBvB'(sin @B') which gives us
mAvA'(sin 30) = mBvB'(sin @B')
Start plugging in numbers...
The x direction...(.12)(2.8) = (.12)(2.1)(cos 30) + .14(vBX)
vBX = .841 m/s
The y direction
(.12)(2.1)(sin 30) = .14(vBY)
vBY = .9 m/s downward
The net for vB'is from the Pythagorean Theorem
v2 = (.9)2 + (.841)2
v = 1.23 m/s
The direction is from the angent function
tan(angle) = .9/.841
angle = 46.9o opposite of the angle for vA'
Thus vB' = 1.23 m/s at an angle of 46.9o (opposite if vA')
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