Answer:
a) v2 = 0 m/s
After the head-on collision:
Particle 1 moves at a velocity of 2 m/s in the same direction as before the collision.
Particle 2 remains at rest.
b) Particle 1 only moves along the x-axis with a velocity of approximately 15.53 m/s, 50° deflected from its original direction of motion.
Particle 2 remains at rest, as in the head-on collision scenario.
Step-by-step explanation:
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy for elastic collisions. In an elastic collision, both momentum and kinetic energy are conserved.
Let's first find the velocity of each particle after the collision in the head-on scenario:
(a) Head-on collision:
Let:
Mass of particle 1 (moving at 2 m/s) = m1 = 5 kg
Mass of particle 2 (initially at rest) = m2 = 8 kg
Velocity of particle 1 before the collision = u1 = 2 m/s
Velocity of particle 2 before the collision = u2 = 0 m/s (at rest)
After the collision, let:
Velocity of particle 1 = v1
Velocity of particle 2 = v2
Using the conservation of momentum:
m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2
5 kg * 2 m/s + 8 kg * 0 m/s = 5 kg * v1 + 8 kg * v2
10 kg + 0 kg = 5 kg * v1 + 8 kg * v2
10 kg = 5 kg * v1 + 8 kg * v2...........(Equation 1)
Using the conservation of kinetic energy:
(1/2) * m1 * u1^2 + (1/2) * m2 * u2^2 = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
(1/2) * 5 kg * (2 m/s)^2 + (1/2) * 8 kg * (0 m/s)^2 = (1/2) * 5 kg * v1^2 + (1/2) * 8 kg * v2^2
(1/2) * 5 kg * 4 m^2/s^2 + 0 = (1/2) * 5 kg * v1^2 + 0
10 kg m^2/s^2 = (1/2) * 5 kg * v1^2
20 m^2/s^2 = 5 kg * v1^2
v1^2 = 20 m^2/s^2 / 5 kg
v1^2 = 4 m^2/s^2
v1 = √(4 m^2/s^2)
v1 = 2 m/s...........(Equation 2)
Now, we have two equations (Equation 1 and Equation 2) to solve for v1 and v2:
10 kg = 5 kg * v1 + 8 kg * v2...........(Equation 1)
v1 = 2 m/s...........(Equation 2)
Substitute Equation 2 into Equation 1:
10 kg = 5 kg * 2 m/s + 8 kg * v2
10 kg = 10 kg + 8 kg * v2
8 kg * v2 = 10 kg - 10 kg
8 kg * v2 = 0 kg
v2 = 0 kg / 8 kg
v2 = 0 m/s
So, after the head-on collision:
Particle 1 moves at a velocity of 2 m/s in the same direction as before the collision.
Particle 2 remains at rest.
(b) Deflected collision at 50°:
In this case, we'll need to break down the initial velocity of particle 1 into its components.
Let:
θ = 50° (angle of deflection)
u1 = 2 m/s (initial velocity of particle 1)
v1 = velocity of particle 1 after the collision
v1x = velocity of particle 1 along the original direction
v1y = velocity of particle 1 perpendicular to the original direction
v1x = v1 * cos(θ)
v1y = v1 * sin(θ)
Using the conservation of momentum in the x-direction:
m1 * u1 = m1 * v1x
5 kg * 2 m/s = 5 kg * v1 * cos(50°)
v1 * cos(50°) = 10 m/s
v1 = 10 m/s / cos(50°)
v1 ≈ 10 m/s / 0.643
v1 ≈ 15.53 m/s
Using the conservation of momentum in the y-direction:
m1 * 0 m/s = m1 * v1y
0 kg = 5 kg * v1 * sin(50°)
v1 * sin(50°) = 0 m/s
v1 = 0 m/s
The above equation tells us that v1 in the y-direction is 0, which means particle 1 does not change its direction in the y-axis. Therefore, particle 1 only moves along the x-axis with a velocity of approximately 15.53 m/s, 50° deflected from its original direction of motion.
Particle 2 remains at rest, as in the head-on collision scenario.