1. find the center of mass (relative to (0,0)) for the spheres with the following masses and locations: m1 = 5kg, (1,1) m2 = 10 kg, (3,1) m3 = 15kg, (1,6).
2. find the center of mass of the following particles (drawn large so they can be seen):
3. an old go - kart with a mass of 300 kg is traveling in a straight line at 80 m/s. it is followed by a 4 - wheeler with mass of 200 kg moving at 60 m/s. how fast is the center of mass moving?
4. a 1500 kg vw is heading 40 m/s in a straight line. a 4000 kg cadillac is heading directly for it at 60 m/s. find the velocity (magnitude and direction) of the center of mass.
5. a 1500 kg car is at rest. at the instant it starts to move (with an acceleration of 3.5m/s²), a truck (m = 3000kg) traveling at a constant speed of 12 m/s passes it. at t = 3 seconds:
a. how far is the center of mass of the vehicles, relative to the starting point of the car?
b. what is the speed of the center of mass of the vehicles?
6. a rock, of mass m, is dropped at t = 0 seconds. two seconds later a stone, of mass 2m, is dropped. at t = 3 seconds (assume neither hits the ground):
a. what is the center of mass of the rock and stone relative to the drop point?
b. how fast is the center of mass going at this time?
7. calculate the vcm before the collision and then calculate the vcm after the collision. (show all work for this problem)
before collision: vo = 5 m/s vo = 0
after collision vf = 1 m/s vf =?

Question

1. find the center of mass (relative to (0,0)) for the spheres with the following masses and locations: m1 = 5kg, (1,1) m2 = 10 kg, (3,1) m3 = 15kg, (1,6).
2. find the center of mass of the following particles (drawn large so they can be seen):
3. an old go - kart with a mass of 300 kg is traveling in a straight line at 80 m/s. it is followed by a 4 - wheeler with mass of 200 kg moving at 60 m/s. how fast is the center of mass moving?
4. a 1500 kg vw is heading 40 m/s in a straight line. a 4000 kg cadillac is heading directly for it at 60 m/s. find the velocity (magnitude and direction) of the center of mass.
5. a 1500 kg car is at rest. at the instant it starts to move (with an acceleration of 3.5m/s²), a truck (m = 3000kg) traveling at a constant speed of 12 m/s passes it. at t = 3 seconds:
 a. how far is the center of mass of the vehicles, relative to the starting point of the car?
 b. what is the speed of the center of mass of the vehicles?
6. a rock, of mass m, is dropped at t = 0 seconds. two seconds later a stone, of mass 2m, is dropped. at t = 3 seconds (assume neither hits the ground):
 a. what is the center of mass of the rock and stone relative to the drop point?
 b. how fast is the center of mass going at this time?
7. calculate the vcm before the collision and then calculate the vcm after the collision. (show all work for this problem)
before collision: vo = 5 m/s vo = 0
after collision vf = 1 m/s vf =?

Accepted Answer

1.
# Explanation:
## Step1: Recall center - of - mass formula for x - coordinate
The formula for the x - coordinate of the center of mass $x_{cm}=\frac{\sum_{i = 1}^{n}m_ix_i}{\sum_{i = 1}^{n}m_i}$. Here, $m_1 = 5kg$, $x_1=1$, $m_2 = 10kg$, $x_2 = 3$, $m_3=15kg$, $x_3 = 1$, and $\sum_{i=1}^{3}m_i=m_1 + m_2+m_3=5 + 10+15=30kg$. Then $x_{cm}=\frac{5	imes1+10	imes3 + 15	imes1}{30}=\frac{5 + 30+15}{30}=\frac{50}{30}=\frac{5}{3}$.
## Step2: Recall center - of - mass formula for y - coordinate
The formula for the y - coordinate of the center of mass $y_{cm}=\frac{\sum_{i = 1}^{n}m_iy_i}{\sum_{i = 1}^{n}m_i}$. Here, $y_1 = 1$, $y_2 = 1$, $y_3=6$. Then $y_{cm}=\frac{5	imes1+10	imes1+15	imes6}{30}=\frac{5 + 10+90}{30}=\frac{105}{30}=\frac{7}{2}$.
# Answer:
$(\frac{5}{3},\frac{7}{2})$

2.
# Explanation:
## Step1: Calculate x - coordinate of center of mass
Let the 20 - kg particle be at $x = 0$, the 5 - kg particle be at $x = 6m$, and the 10 - kg particle be at $x=6 + 2=8m$. $\sum_{i=1}^{3}m_i=20 + 5+10=35kg$. $x_{cm}=\frac{20	imes0+5	imes6+10	imes8}{35}=\frac{0 + 30+80}{35}=\frac{110}{35}=\frac{22}{7}m$.
## Step2: Calculate y - coordinate of center of mass
Let the 20 - kg particle be at $y = 0$, the 5 - kg particle be at $y = 0$, and the 10 - kg particle be at $y = 3m$. $y_{cm}=\frac{20	imes0+5	imes0+10	imes3}{35}=\frac{30}{35}=\frac{6}{7}m$.
# Answer:
$(\frac{22}{7}m,\frac{6}{7}m)$

3.
# Explanation:
## Step1: Recall the formula for the velocity of the center of mass
The formula for the velocity of the center of mass $v_{cm}=\frac{m_1v_1+m_2v_2}{m_1 + m_2}$, where $m_1 = 300kg$, $v_1 = 80m/s$, $m_2=200kg$, $v_2 = 60m/s$, and $m_1 + m_2=300 + 200=500kg$. Then $v_{cm}=\frac{300	imes80+200	imes60}{500}=\frac{24000+12000}{500}=\frac{36000}{500}=72m/s$.
# Answer:
$72m/s$

4.
# Explanation:
## Step1: Define the positive direction
Let the direction of the VW be the positive direction. Then $m_1 = 1500kg$, $v_1 = 40m/s$, $m_2 = 4000kg$, $v_2=-60m/s$, and $\sum_{i=1}^{2}m_i=1500 + 4000=5500kg$.
## Step2: Calculate the velocity of the center of mass
$v_{cm}=\frac{m_1v_1+m_2v_2}{m_1 + m_2}=\frac{1500	imes40+4000	imes(-60)}{5500}=\frac{60000-240000}{5500}=\frac{- 180000}{5500}=-\frac{360}{11}\approx - 32.73m/s$. The negative sign indicates the direction is the same as the direction of the Cadillac.
# Answer:
Magnitude: $\frac{360}{11}m/s\approx32.73m/s$, Direction: The same as the direction of the Cadillac

5.
# Explanation:
## Step1a: Calculate the position of the car and the truck at $t = 3s$
For the car, using $x_1=v_0t+\frac{1}{2}at^{2}$, with $v_0 = 0$, $a = 3.5m/s^{2}$, $t = 3s$, $x_1=\frac{1}{2}	imes3.5	imes3^{2}=\frac{1}{2}	imes3.5	imes9 = 15.75m$. For the truck, using $x_2=v_2t$, with $v_2 = 12m/s$, $t = 3s$, $x_2=12	imes3 = 36m$. $m_1 = 1500kg$, $m_2 = 3000kg$, $\sum_{i=1}^{2}m_i=1500+3000 = 4500kg$. $x_{cm}=\frac{m_1x_1+m_2x_2}{m_1 + m_2}=\frac{1500	imes15.75+3000	imes36}{4500}=\frac{23625+108000}{4500}=\frac{131625}{4500}=29.25m$.
## Step1b: Calculate the velocity of the car and the truck at $t = 3s$
For the car, using $v_1=v_0+at$, $v_1=0+3.5	imes3 = 10.5m/s$. For the truck, $v_2 = 12m/s$. $v_{cm}=\frac{m_1v_1+m_2v_2}{m_1 + m_2}=\frac{1500	imes10.5+3000	imes12}{4500}=\frac{15750+36000}{4500}=\frac{51750}{4500}=11.5m/s$.
# Answer:
a. $29.25m$
b. $11.5m/s$

6.
# Explanation:
## Step1a: Calculate the position of the rock and the stone at $t = 3s$
For the rock, using $y_1=\frac{1}{2}gt^{2}$ with $t = 3s$, $y_1=\frac{1}{2}g	imes3^{2}=\frac{9}{2}g$. For the stone, it is dropped 2s later, so $t_2=3 - 2 = 1s$, and $y_2=\frac{1}{2}g	imes1^{2}=\frac{1}{2}g$. $m_1 = M$, $m_2 = 2M$, $\sum_{i=1}^{2}m_i=3M$. $y_{cm}=\frac{M	imes\frac{9}{2}g+2M	imes\frac{1}{2}g}{3M}=\frac{\frac{9}{2}Mg+Mg}{3M}=\frac{\frac{9 + 2}{2}Mg}{3M}=\frac{11}{6}g$.
## Step1b: Calculate the velocity of the rock and the stone at $t = 3s$
For the rock, $v_1=gt=3g$. For the stone, $v_2=gt_2 = g$. $v_{cm}=\frac{M	imes3g+2M	imes g}{3M}=\frac{3Mg+2Mg}{3M}=\frac{5}{3}g$.
# Answer:
a. $\frac{11}{6}g$
b. $\frac{5}{3}g$

7.
# Explanation:
## Step1: Calculate $v_{cm}$ before the collision
Using $v_{cm}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1 + m_2}$, where $m_1 = 4kg$, $v_{1i}=5m/s$, $m_2 = 10kg$, $v_{2i}=0$. $\sum_{i=1}^{2}m_i=4 + 10=14kg$. $v_{cm}=\frac{4	imes5+10	imes0}{14}=\frac{20}{14}=\frac{10}{7}m/s$.
## Step2: Use conservation of momentum to find $v_{2f}$
By conservation of momentum $m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$. Substituting the values: $4	imes5+10	imes0=4	imes1+10	imes v_{2f}$, $20=4 + 10v_{2f}$, $10v_{2f}=16$, $v_{2f}=1.6m/s$.
## Step3: Calculate $v_{cm}$ after the collision
$v_{cm}=\frac{m_1v_{1f}+m_2v_{2f}}{m_1 + m_2}=\frac{4	imes1+10	imes1.6}{14}=\frac{4 + 16}{14}=\frac{20}{14}=\frac{10}{7}m/s$.
# Answer:
Before collision: $\frac{10}{7}m/s$
After collision: $\frac{10}{7}m/s$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall center - of - mass formula for x - coordinate

Step2: Recall center - of - mass formula for y - coordinate

Step1: Calculate x - coordinate of center of mass

Step2: Calculate y - coordinate of center of mass

Step1: Recall the formula for the velocity of the center of mass

Answer: