The eight ball, which has a mass of m = 0.5 kg. is initially moving with a veloc

Question

The eight ball, which has a mass of m = 0.5 kg. is initially moving with a velocity v = 2.5i m s. It the strikes the six ball, which has an identical mass and is initially at rest. After the collision, the eight ball is deflected by an angle of theta = 19 degree and the six ball is deflected by an angle of Phi = 12.5 degree, as shown in the figure. Write an expression for the magnitude of six ball's velocity, in terms of the angles given in the problem and the magnitude of the eight-ball's initial velocity, v. What is the magnitude of the velocity, in meters per second, of the six ball? What is the magnitude of the velocity of the eight ball, in meters per second, after the collision?

Explanation / Answer

a)

We need to use law of conservation of linear momentum,

m1v1i+m2v2i = m1v1f+ m2v2f --------------(1)

Along horizontal axis,

m1v1ix+m2v2i x= m1v1fx+ m2v2fx

m1Vix+m2v2i x= m1v1fx+ m2v2fx

m1V + m2*0 = m1*v1fcos + m2*v2f*cos

m1V = m1*v1fcos + m2*v2f*cos -----------------(2)

Along vertical axis,

m1Viy+m2v2i y= m1v1fy+ m2v2fy

m1*0 + m2*0 = m1*v1fsin + m2*v2f*sin

m1*v1fsin + m2*v2f*sin = 0

m1*v1fsin = -m2*v2f*sin

v2f= (-m1*v1fsin)/( m2*sin ) -----------------(3)

Plug (3) in (2),

m1V = m1*v1fcos + m2*[(-m1*v1fsin)/( m2*sin )]*cos

m1V = m1*v1fcos + [(-m1*v1fsin cos)/( sin )

m1V = v1f*{(m1*cos) + [(-m1*sin cos)/( sin )}

v1f = m1V/{(m1*cos) + [(-m1*sin cos)/( sin )} ------------------(4)

c)

Plugging given data in (4),

v1f = (0.5*2.5)/{(0.5*cos19) + [(-0.5*sin19* cos12.5)/(sin(-12.5 ))}

v1f = 1.035m/s

b)

Plugging above value in (3),

v2f= (-0.5*1.035*sin19)/(0.5*sin(-12.5))

v2f= 1.55m/s

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