The vector position of a 3.40 g particle moving in the xy plane varies in time according to the following equation.
At the same time, the vector position of a 5.05 g particle varies according to the following equation.
For each equation, t is in s and r is in cm. Solve the following when t = 2.80
(b) Find the linear momentum of the system.
g-cm/s
g-cm/s
(c) Find the velocity of the center of mass.
cm/s
(e) Find the net force exerted on the two-particle system.
μN
μN
To solve this problem, we need to find the position vectors of both particles at time t = 2.80 and then calculate the linear momentum and velocity of the center of mass. Finally, we can use the linear momentum to find the net force exerted on the two-particle system.
Let's start with finding the position vectors of the particles at t = 2.80.
For the first particle with a mass of 3.40 g, the position vector is given by:
r1(t) = (2.80t - 1.60)i - (3.60t + 2.00)j
Substituting t = 2.80 into the equation, we get:
r1(2.80) = (2.80(2.80) - 1.60)i - (3.60(2.80) + 2.00)j
= 4.08i - 12.68j
Therefore, the position vector of the first particle at t = 2.80 is 4.08i - 12.68j cm.
For the second particle with a mass of 5.05 g, the position vector is given by:
r2(t) = (0.80t + 3.20)i + (2.80t - 2.40)j
Substituting t = 2.80 into the equation, we get:
r2(2.80) = (0.80(2.80) + 3.20)i + (2.80(2.80) - 2.40)j
= 7.84i + 5.52j
Therefore, the position vector of the second particle at t = 2.80 is 7.84i + 5.52j cm.
Now, let's calculate the linear momentum of the system.
The linear momentum of a particle is given by the product of its mass and velocity. Since we are given the position vectors, we can differentiate them with respect to time to obtain the velocity vectors.
The velocity vector for the first particle is obtained by differentiating r1(t):
v1(t) = d/dt (r1(t)) = d/dt (4.08i - 12.68j) = 0i - 0j = 0
Similarly, the velocity vector for the second particle is obtained by differentiating r2(t):
v2(t) = d/dt (r2(t)) = d/dt (7.84i + 5.52j) = 0i + 0j = 0
Both particles have zero velocity, which means their linear momentum is also zero. Therefore, the linear momentum of the system is 0 g-cm/s.
Next, let's calculate the velocity of the center of mass. The center of mass velocity is given by the total linear momentum divided by the total mass of the system.
The total mass of the system is the sum of the masses of the individual particles:
m_total = m1 + m2 = 3.40 g + 5.05 g = 8.45 g
Since the linear momentum of the system is zero, the velocity of the center of mass is also zero. Therefore, the velocity of the center of mass is 0 cm/s.
Finally, let's calculate the net force exerted on the two-particle system. The net force can be calculated using the equation:
F_net = d(m_total * v_cm)/dt
Since the velocity of the center of mass is zero, the net force is also zero. Therefore, the net force exerted on the two-particle system is 0 μN.