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 = 2.3i 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 = 21degree and the six ball is deflected by an angle of Phi = 19degree, as shown in the figure. Randomized Variables V = 2.3i m/s theta = 21degree Phi = 19degree 33% Part (a) Write an expression for the magnitude of six ball's velocity, in tenns of the angles given in the problem and the magnitude of the eight ball's initial velocity, nu. 33% Part (b) What is the magnitude of the velocity, in meters per second, of the six ball? 33% Part (c) What is the magnitude of the velocity of the eight ball, in meters per second, after the collision?

Explanation / Answer

From law of conservation of momentum along vertical direction of ball 8 is

m1v1sin-m2v2sin = 0

v2 = m1v1sin/m2sin

v2 = 0.5xv1 x sin (21o) / (0.5 x sin (19O)

v2= 0.179v1 / 0.163 = 1.059 v1

apply law of conservation of momentum along the horizontal direction

m1u1 = m1v1cos+ m2v2cos

= m1v1cos+ m2 (1.059v1) cos

=v1 (m1cos+1.059 m2cos )

u1= v1 (cos + 1.059 (m2/m1)cos

2.3= v1 (cos 21 + 1.059 x1x cos 19) masses are identical so m2/m1 =1

2.3 =v1 (0.93 + 1.059 x0.946)

2.3/1.932 =v1

v1=1.191m/s

velocity of ball 6 is 1.191m/s

after collision ball 8 movies with speed of 1.059x1.191 = 1.261m/s

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