The figure gives an overhead view of the path taken by a 016

The figure gives an overhead view of the path taken by a 0.168 kg cue ball as it bounces from a rail of a pool table. The ball\'s initial speed is 1.42 m/s, and the angle ?1 is 64.5°. The bounce reverses the y component of the ball\'s velocity but does not alter the x component. What are (a) angle ?2 and (b) the magnitude of the change in the ball\'s linear momentum? (The fact that the ball rolls is irrelevant to the problem.)

Solution

given that

mass m = 0.168 kg

initial speed v = 1.42 m/s

theta 1 = 64.5 degree

part (a)

As given that

the x component of velocity is unchanged and the y component is reversed

so ,theta2 is equal to theta1, but in the opposite direction

so theta 2 = -64.5 degree

part (b)

initial momentum is

p1 = (m * v * sin(theta1)) i + (m * v * cos(theta1)) j

momentum after collision

p2 = (m * v * sin(theta1)) i - (m * v * cos(theta1)) j

so change in momentum is

delta p = p2 - p1

delta p = 0i+ (-2 * m * v * cos(theta 1) j )

magnitude of delta p = sqrt ( (delta p)^2 )

delta p = sqrt (-2*m*v*cos(theta1))^2

delta p = 2*m*v*cos(theta1)

deltap = 2*0.168*1.42*cos(64.5)

delta p = 0.205 kg*m/s