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.

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## 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...