The figure gives an overhead view of the path taken by a cue ball with mass m as
ID: 1499236 • Letter: T
Question
The figure gives an overhead view of the path taken by a cue ball with mass m as it bounces from a rail of a pool table. The ball's initial speed is vi and the angle of impact is 1. 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 change in the ball's linear momentum in unit-vector notation? (The fact that the ball rolls is irrelevant to the problem.) State your answers in terms of the given variables (m, vi, 1).
I got Theta 1 = Theta 2 for part A.) but can not find the right answer for part B.)
Part B.) is given in_________ i(hat) +_____ j(hat)
Explanation / Answer
I think that that part of the motion is already clear to you.
In order to properly describe the direction of the velocity vector,
it is best to look at the components of the vector along the
coordinate axes. Since vector V1, the velocity makes the angle theta1 with respect
to the x axis, then its x component V1x and y component
V1y must be given by
(3) ..... V1x = +V1*cos(theta1)
positive because it is pointing to the right in the (+x) direction
(4) ..... V1y = -V1*sin(theta1)
negative because it is pointing downward in the (-y) direction.
Now, in bouncing back with the velocity vector V2 making the
angle theta2 with respect to the horizontal, we find that the x
and y components are
(5) ..... V2x = +V2*cos(theta2)
positive because it is pointing to the right in the (+x) direction
(6) ..... V2y = +V2*sin(theta1)
positive because it is pointing upward in the (+y) direction.
Now, since the bounce reverses the y component of the ball's
velocity but does not alter the x component, then, we find
from (2), (3), and (5) that V1x = V2x , so that
(7) ..... +V1*cos(theta1) = +V2*cos(theta2) = +V2*cos(theta1)
which gives V1 = V2 = Vo
Rewriting the components of the velocity vector V1, we get
(8) ..... V1x = +Vo*cos(theta1)
(9) ..... V1y = - Vo*sin(theta1)
Rewriting the components of the velocity vector V2, we get
(10) .... V2x = +Vo*cos(theta1)
(11) .... V2y = + Vo*sin(theta1)
Now, according to (1), we get the following components for the
linear momentum P1 before bouncing off :
(12) ..... P1x = M * V1x = +M * Vo*cos(theta1)
(13) ..... P1y = M * V1y = - M * Vo*sin(theta1)
Similarly, we get the following components for the
linear momentum P2 after bouncing off :
(14) ..... P2x = M * V2x = +M * Vo*cos(theta1)
(15) ..... P2y = M * V2y = + M * Vo*sin(theta1)
According to (12) and (14), the change in the ball's linear momentum
dPx along the x-axis is
(16) .... dPx = P2x - P1x = 0
According to (13) and (15), the change in the ball's linear momentum
dPy along the y-axis is
(16) .... dPy = P2y - P1y = +2 * M * Vo*sin(theta1)
Notice that since dPx = 0 and dPy is positive, the change in linear momentum
is pointing upward in (+y) direction.
So, answer is : 0 i(hat) +2MV0 sin(theta1) j(hat)