So... I REALLY can't find any help online nor any help from anyone. So, I was wondering if you could help me solve this problem.
Find the center of mass of the three mass system shown in Figure 7-38 given that m1 = 1.50 kg and m2 = 1.25 kg. Specify relative to the center of the left hand (1.00 kg) mass.
Figure : http://i154.photobucket.com/albums/s266/MszxCyndii...
Answer is in : __________________ m
A ball of mass 0.640 kg moving east (+x direction) with a speed of 3.10 m/s collides head-on with a 0.820 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?
ball originally at rest 2.717 m/s EAST
ball originally moving east _________________ m/s East or West?
Thank you.
Thumbs up to anyone that helps.
Answers & Comments
Verified answer
a. Find com.
com = sum m*d / total mass
com = m0d0 + m1d1 +m2d2 / (m0 + m1 + m2)
m0= 1 kg
m1= 1.5 kg
m2 = 1.25 kg
d0 = 0
d1 = .5 m
d2 = .75 m
com = (1 kg * 0 + 1.5kg *.5m + 1.25kg * .75m)/ (1kg + 1.5 kg + 1.25 kg)
com = (0 + .75 kg m + .9375 kg m)/ (3.75 kg)
com = 1.6875 kg m / 3.75 kg
com = .45 m
b. 0.38219 West
(a). ω² = k/m = 300/4 ω = 8.66 rad/s y = A cos (ω t) y' = -Aω sin (ω t) y'/y = -ω tan (ω t) 0.55/0.02 = -8.66 tan (8.66 t) t = 0.2166 s y = A cos (ω t) 0.02 = A cos (8.66 * 0.216) A = 1.876 m (b). y' = -Aω sin (ω t) y' max = Aω = 1.876 (8.66) y' max = 16.25 m/s please correct me if i am wrong...