pool cue vector
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Linear Momentum of a System of Particles (Stuck on certain part)?
The path taken by a 0.165 kg cue ball as it bounces from the rail of a pool table. The ball’s initial speed is 1.50 m/s, and the angle θ1 = 30.0°. The bounce reverses the y component of the ball’s velocity but does not alter the x component.
What is the change in the ball’s linear momentum in unit-vector notation? (The fact that the ball rolls is irrelevant to the question.)
-Answer should be in terms of
() i kg·m/s + () j kg·m/s
let’s call the direction parallel to the rail the x direction, and the perpendicular direction the y direction; let’s call away from the rail the +y direction
momentum = mv where v is a vector, so we can write momentum in components:
momentum = (mv)i + (mv)j where x and y denote the components of the momentum in each direction
the initial x momentum is 0.165(1.5cos30)
the final x momentum is the same, so there is no change
the final y momentum is 0.165kgx1.5 sin30 m/s = +0.12j
the initial y momentum is -0.12y
therefore the change in momentum is 0i + (0.12-(-0.12))j
change in momentum = 0i +0.24j
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