Motion of a System of Particles Problems and Solutions

 Problem#1

A 2.00-kg particle has a velocity (2.00i – 3.00j) m/s, and a 3.00-kg particle has a velocity (1.00i + 6.00j) m/s. Find (a) the velocity of the center of mass and (b) the total momentum of the system.

Answer:
(a) the velocity of the center of mass is

v_CM = ∑m_iv_i/M = (m₁v₁ + m₂v₂)/M

= [(2.00kg)(2.00i – 3.00j) m/s + (3.00kg)(1.00i + 6.00j) m/s]/(5.00kg)

v_CM = (1.40i + 2.40j) m/s

(b) the total momentum of the system is

p = Mv = (5.00kg)(1.40i + 2.40j) m/s = (7.00i + 12.0j) kg.m/s

Problem#2
Consider a system of two particles in the xy plane: m₁ = 2.00 kg is at the location r₁ = (1.00i + 2.00j) m and has a velocity of (3.00i + 0.500j) m/s; m₂ = 3.00 kg is at r₂ = (–4.00i – 3.00j) m and has velocity (3.00i – 2.00j) m/s. (a) Plot these particles on a grid or graph paper. Draw their position vectors and show their velocities. (b) Find the position of the center of mass of the system and mark
it on the grid. (c) Determine the velocity of the center of mass and also show it on the diagram. (d) What is the total linear momentum of the system?

Answer:
(a)

(b) Using the definition of the position vector at the center of mass,

r_CM = (m₁r₁ + m₂r₂)/(m₁ + m₂)

r_CM = [(2.00kg)(1.00m, 2.00m) + (3.00kg)(–4.00m, –3.00m)]/(2.00 kg + 3.00 kg)

r_CM = (–2.00i – 1.00j)m

(c) The velocity of the center of mass is

v_CM = (m₁v₁ + m₂v₂)/(m₁ + m₂)

v_CM = [(2.00kg)(3.00m m/s, 0.500 m/s) + (3.00kg)(3.00m/s, –2.00m/s)]/(2.00 kg + 3.00 kg)

v_CM = (3.00i – 1.00j)m/s

(d) The total linear momentum of the system can be calculated as

P = Mv_CM

P = (5.00 kg)(3.00i – 1.00j)m/s

P = (15.0i – 5.00j)m/s

Problem#3
Romeo (77.0 kg) entertains Juliet (55.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.70 m away from Juliet, who is in the front of the boat. After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo’s cheek. How far does the 80.0-kg boat move toward the shore it is facing?

Answer:

Let x = distance from shore to center of boat

L = length of boat

x′ = distance boat moves as Juliet moves toward Romeo

The center of mass stays fixed.

Before:

x_CM = [M_Bx + M_J(x – L/2) + M­_R(x + L/2)]/(M_B + M_J + M_R)

After:
x_CM = [M_B(x – x’) + M_J(x + L/2 – x’) + M­_R(x + L/2 – x’)]/(M_B + M_J + M_R)

There is no external force, hence COM will not move. Then

x_CM (BEFORE) = x_CM(AFTER)

[M_Bx + M_J(x – L/2) + M­_R(x + L/2)] = [M_B(x – x’) + M_J(x + L/2 – x’) + M­_R(x + L/2 – x’)]
M_Bx + M_Jx – M_JL/2 + M­_Rx + M­_RL/2 = M_Bx – M_Bx’ + M_Jx + M_JL/2 – M_Jx’ + M­_Rx + M­_RL/2 – M­_Rx’

 – M_JL/2 = – M_Bx’ + M_JL/2 – M_Jx’ – M­_Rx’

(M_B + M_J + M_R)x’ = M_JL

(80.0 kg + 55.0 kg + 77.0 kg)x’ = (55.0kg)(2.70 m)

x' = 0.700 m

Problem#4
A ball of mass 0.200 kg has a velocity of 150i m/s; a ball of mass 0.300 kg has a velocity of –0.400i m/s. They meet in a head-on elastic collision. (a) Find their velocities after the collision. (b) Find the velocity of their center of mass before and after the collision.

Answer:
(a) Conservation of momentum for the two-ball system gives us:

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

(0.200 kg)(1.50 m/s) + (0.300 kg)(–0.400 m/s) = (0.200 kg)v_1f + (0.300 kg)v_2f

Relative velocity equation:

v_2f – v_1f = 1.90 m/s
then,
(0.300 kg.m/s – 0.120 kg.m/s) = (0.200 kg)v_1f + (0.300 kg)(v_1f + 1.90 m/s)

–0.390 kg.m/s = 0.500kgv_if

v_1f = –0.780 m/s or v_1f = –0.780i m/s

and v_2f = 1.90 m/s – 0.780 m/s = 1.12 m/s or v_2f = 1.12i m/s

(b) the velocity of their center of mass before and after the collision.

Before:
v_CM = (m₁v_1i + m₂v_2i)/(m₁ + m₂)

v_CM = [(0.20kg)(1.50i m/s) + (0.30kg)(–0.40i m/s)]/(0.20 kg + 0.30 kg)

v_CM = (0.360 m/s)i

Afterwards, the center of mass must move at the same velocity, as momentum of the system
is conserved.  

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