Question

Billiard ball A of mass mA = 0.115 kg moving with speed vA = 2.80 m/s strikes ball B , initially at rest, of mass mB = 0.144 kg . As a result of the collision, ball A is deflected off at an angle of θ′A = 30.0∘ with a speed v′A = 2.10 m/s , and ball B moves with a speed v′B at an angle of θ′B to original direction of motion of ball A.

Part A

Taking the x axis to be the original direction of motion of ball A , choose the correct equation expressing the conservation of momentum for the components in the x direction.

Taking the x axis to be the original direction of motion of ball A, choose the correct equation expressing the conservation of momentum for the components in the x direction.

	0=mAv′Asinθ′A−mBv′Bsinθ′B
	mAvA=mAv′Acosθ′A+mBv′Bcosθ′B
	mAvA=mAv′Acosθ′A−mBv′Bsinθ′B
	0=(mAvA+mBv′B)sinθ′B Part B Taking the x axis to be the original direction of motion of ball A , choose the correct equation expressing the conservation of momentum for the components in the y direction. mAvA=mAv′Acosθ′A−mBv′Bsinθ′B 0=(mAvA+mBv′B)sinθ′B 0=mAv′Asinθ′A−mBv′Bsinθ′B mAvA=mAv′Acosθ′A+mBv′Bcosθ′B Part C Solve these equations for the angle, θ′B , of ball B after the collision. Do not assume the collision is elastic. Part D Solve these equations for the speed, v′B , of ball B after the collision. Do not assume the collision is elastic.

Answer 1

mAvA=mAv′Acosθ′A+mBv′Bcosθ′B

part b )

0=mAv′Asinθ′A−mBv′Bsinθ′B

because initally there is no y component

ma*vai = ma*vaf + mb*vbf

vbf = ma/mb*vai - ma/mb*vaf

vbf = ma/mb*(vai - vaf)

vbf = ma/mb*(vai - vaf)

vai = 2.8i + 0j m/s

vaf = 2.10 m/s * (cos(30)i + sin(30)j) = 1.819i + 1.05j m/s

vbf = 0.115 kg / 0.144 kg * ((2.8i + 0j) m/s - (1.819i + 1.05j) m/s)

vbf = 0.115/0.144*(0.981i - 1.05j m/s)

vbf = 0.783i - 0.838 j

part c )

theta = tan^-1(0.838/0.783) = 46.94 degree = 47 degree

part d )

vb = sqrt(vx^2 + vy^2)

vb = sqrt(.783^2 + .838^2)

vb = 1.147 m/s