(II) Billiard ball A of mass mA = 0.120 kg moving with speed vA = 2.80 m/s strik

Question

(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 v'A = 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, v'B, and angle, theta'B, of ball B after the collision. Do not assume the collision is elastic.

Explanation / Answer

mA = 0.12

vAx = 2.8m/s

vAy = 0

after collision

vAx' = vA'*cos30 = 2.1*cos30 = 1.82 m/s

vAy' = vA'*sin30 = 1.05 m/s

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mB = 0.14 kg

before collision

vBx = 0    vBy = 0

after collison

vBx' = vB'*costhetaB'

VBy' = VB'*sinthetaB'

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according to aonservation of momentum the total momentum remains same before and after the collision

along x direction

Px = Px'

mA*vAx + mB*vBx = mA*VA'x + mB*vB'x

along y direction

mA*vAy + mB*vBy = mA*VA'y + mB*vB'y

++++++++++++++++++++

part(b)

along y direction

mA*0 + mB*0 = (0.12*1.05)+(0.14*vBy')

VBy' = -0.9 m/s

along x

(0.12*2.8)+0 = (0.12*1.82)+(0.14*vBy')

vBx' = 0.84 m/s <---answer

speed vB' = sqrt(VBx'^2 + VBy'^2)

V'B = 1.23 m/s

direction

tanthetaB' = VBy'/VBx'

thetaB' = 46.97 degrees with the x axis in clock wise direction

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