At the instant a 2.0-kg particle has a velocity of 4.0 m/s in the positive x direction, a 3.0-kg particle has a velocity of 5.0 m/s in the positive y direction. What is the speed of the center of mass of the two-particle system?
find total momentum, the combined momentum is the same (no external forces) For example a man on one lassos the other and pulls them together. There was no external force, so momentum of combined mass is the same as the original two.
x direction:
2 * 4 = 8
y direction
3*5 = 15
total momentum = sqrt (64+225) = sqrt (289) = 17
so total mass of 5 kg is moving at speed s
5 s = 17
s = 17/5
Well, it seems like these particles are on their way to star in their own action-packed movie! To find the speed of the center of mass of the two-particle system, we can use the formula:
Velocity of center of mass = (m1 * v1 + m2 * v2) / (m1 + m2)
So let's plug in the values, shall we?
m1 = 2.0 kg, v1 = 4.0 m/s
m2 = 3.0 kg, v2 = 5.0 m/s
(2.0 kg * 4.0 m/s + 3.0 kg * 5.0 m/s) / (2.0 kg + 3.0 kg)
Now, let me grab my calculator... *beep boop beep*... and we get:
(8.0 kg*m/s + 15.0 kg*m/s) / 5.0 kg
23.0 kg*m/s / 5.0 kg
*drumroll please*...
The speed of the center of mass is 4.6 m/s!
So, it looks like the center of mass of this dynamic duo is moving at quite a decent speed. They're definitely chasing their dreams together!
To find the speed of the center of mass of the two-particle system, we need to determine the total momentum of the system and divide it by the total mass.
The momentum of each particle is given by:
p1 = m1 * v1
p2 = m2 * v2
Where:
m1 and m2 are the masses of the particles
v1 and v2 are the velocities of the particles
Given:
m1 = 2.0 kg
v1 = 4.0 m/s
m2 = 3.0 kg
v2 = 5.0 m/s
Let's calculate the momentum of each particle:
p1 = (2.0 kg) * (4.0 m/s) = 8.0 kg m/s
p2 = (3.0 kg) * (5.0 m/s) = 15.0 kg m/s
The total momentum of the system, P total, is given by the vector sum of the individual momenta:
P total = p1 + p2
P total = 8.0 kg m/s + 15.0 kg m/s
P total = 23.0 kg m/s
Next, we need to calculate the total mass of the system:
m total = m1 + m2
m total = 2.0 kg + 3.0 kg
m total = 5.0 kg
Finally, we can calculate the speed of the center of mass, V cm, by dividing the total momentum by the total mass:
V cm = (P total) / (m total)
V cm = 23.0 kg m/s / 5.0 kg
V cm = 4.6 m/s
Therefore, the speed of the center of mass of the two-particle system is 4.6 m/s.
To find the speed of the center of mass of a two-particle system, we need to calculate the velocity components of the individual particles and then find the resultant velocity.
Let's start by finding the velocity components of each particle.
For the 2.0-kg particle with a velocity of 4.0 m/s in the positive x direction, the velocity vector would be (4.0 m/s, 0 m/s).
For the 3.0-kg particle with a velocity of 5.0 m/s in the positive y direction, the velocity vector would be (0 m/s, 5.0 m/s).
Now, let's find the center of mass velocity.
Since the center of mass is the average of the individual particle velocities, we can calculate it by taking the weighted average using the masses of the particles.
Let's denote the center of mass velocity vector as (Vcmx, Vcmy). We can use the following formula:
Vcmx = (m1 * V1x + m2 * V2x) / (m1 + m2)
Vcmy = (m1 * V1y + m2 * V2y) / (m1 + m2)
Substituting the given values:
m1 = 2.0 kg (mass of the first particle)
V1x = 4.0 m/s (x-component velocity of the first particle)
V1y = 0 m/s (y-component velocity of the first particle)
m2 = 3.0 kg (mass of the second particle)
V2x = 0 m/s (x-component velocity of the second particle)
V2y = 5.0 m/s (y-component velocity of the second particle)
Vcmx = (2.0 kg * 4.0 m/s + 3.0 kg * 0 m/s) / (2.0 kg + 3.0 kg)
= 8.0 kg m/s / 5.0 kg
= 1.6 m/s
Vcmy = (2.0 kg * 0 m/s + 3.0 kg * 5.0 m/s) / (2.0 kg + 3.0 kg)
= 15.0 kg m/s / 5.0 kg
= 3.0 m/s
Therefore, the velocity of the center of mass is (1.6 m/s, 3.0 m/s) when rounded to the appropriate number of significant figures.
Now, to calculate the speed of the center of mass, we can use the Pythagorean theorem:
Speed of the center of mass = √(Vcmx^2 + Vcmy^2)
= √((1.6 m/s)^2 + (3.0 m/s)^2)
= √(2.56 m^2/s^2 + 9.0 m^2/s^2)
= √(11.56 m^2/s^2)
= 3.4 m/s (rounded to the appropriate number of significant figures)
Therefore, the speed of the center of mass of the two-particle system is approximately 3.4 m/s.