weekly physics?

Question

three uniform thin rods, each of length L = 24 cm, form an inverted U. The vertical rods each have mass of 14 g; the horizontal rod has mass of 32 g.


Where is the center of mass of the assembly?
(x, y) = ( , ) cm



a 0.165 kg cue ball bounces from the rail of a pool table. The ball's initial speed is 2.30 m/s, and the angle with the vertical = 30.0°. The bounce reverses the y component of the ball's velocity but does not alter the x component.

(b) 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.)
____i kgm/s + ____j kgm/s

Answer 1

Let me see. you have three rods, right ?

So, if you analize it, you must look for the x and y position for the center of mass, so :

for the first rod, the x and y position for its center of mass, will be : x,y = 0,12, because the first one is vertical

the second horizontal rod, will have its center of mass in :
x,y = 12,24

The third rod, the other vertical rod, will have its center of mass in:

x,y = 24,12, why x = 24 ?, because, the horizontal rod is 24 cm

then, the x for the center of mass will be :

x = 14*0 + 32*12 + 14*24 / 14 +14 + 32

x = 12 cm

and y for the center of mass will be :

y = 14*12 + 32*24 + 14*12 / 14 +14 +32

y = 18.4 cm

b) If the bounces reverse the y component :

the x component for its speed will be the same : 2,3*cos60

Vx = 1.15 m/s

then the linear momentum fot Vx = 1.15*0.165 = 0.18 kgm/s

for Vy = -2.3sen60 = -1.99 m/s

the linear momentum for Vy = -0.33 kgm/s

0.18 i - 0.33 j

Answer 2

To find the center of mass, draw the picture, add up all the moments (mass times distance to the individual centers of mass of the rods, which are obviously in their centers) in the x and y directions. Divide the sum of the moments by the sum of the masses, and that's your center of mass.

The problem states that the x velocity (and therefore the momentum) are unchanged by the bounce.

Had to edit this next [part because I misread.

The y velocity is just the velocity times the cosine of that angle (because it's measured from the normal, it's not sine). The ball bounces from negative to positive, so the momentum change is twice the mass times that velocity component.

Got it?