Question

with a velocity v = 4.6i 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 θ = 26° and the six ball is deflected by an angle of Φ = 35°, as shown in the figure.

Part (a) 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.

Part (b) What is the magnitude of the velocity, in meters per second, of the six ball?

Part (c) What is the magnitude of the velocity of the eight ball, in meters per second, after the collision?

Answer 1

m1 = m = 0.5 kg               m2 = m = 0.5 kg

befor collision

v1 = 4.6i                   v2 = 0

after collision

v1'x = v1'*cos26                  v2'x = v2'*cos35

v1'y = v1'*sin26                  v2'y = v2'*sin35

from momentum conservation

along y

Piy = Pfy

0 = m1*v1'y + m2*v2'y

v1'y = v2'y

v1'*sin26 = v2'*sin35

v1' = v2'*sin35/sin26......(1)

along x axis

Pix = Pfx

m1*v = m1*v1'x + m2*v2'x

v = v1'*cos26 + v2'*cos35.......(2)

using 1 in 2

v = v2'*sin35*cos26/sin26 + v2'*cos35

v*sin26 = v2'*sin35*cos26 + v2' *cos35*sin26

v*sin26 = v2'*sin(35+26)

v*sin26 = v2'*sin61       <------answer

-------

part(b)

v2' = 4.6*sin26/sin61 = 2.3 m/s <------answer

part(c)

v1' = v2'*sin35/sin26

v1' = 3 m/s <<------answer