particle 1 of mass m1 = 0.29 kg slides rightward along an x axis on a frictionless floor with a speed of 1.8 m/s. When it reaches x = 0, it undergoes a one-dimensional elastic collision with stationary particle 2 of mass m2 = 0.40 kg. When particle 2 then reaches a wall at xw = 70 cm, it bounces from it with no loss of speed. At what position on the x axis does particle 2 then collide with particle 1?
To find the position on the x-axis where particle 2 collides with particle 1 after bouncing off the wall, we need to use the principles of conservation of momentum and conservation of kinetic energy.
Let's break down the problem and find the velocities of the particles before and after the collision:
Given:
- Mass of particle 1, m1 = 0.29 kg
- Velocity of particle 1 before collision, v1 = 1.8 m/s
- Mass of particle 2, m2 = 0.40 kg
- Velocity of particle 2 before collision, v2 = 0 m/s (since it's stationary)
Using the conservation of momentum, we can write:
m1*v1i + m2*v2i = m1*v1f + m2*v2f
Here, v1i and v2i represent the initial velocities of particles 1 and 2, and v1f and v2f represent their final velocities after the collision.
Since particle 1 is moving in the positive x-direction and particle 2 is stationary, we can rewrite the equation as:
m1*v1i = m1*v1f + m2*v2f (equation 1)
Next, we'll use the conservation of kinetic energy to relate the velocities of the particles before and after the collision:
(1/2)*m1*(v1i)^2 + 0 = (1/2)*m1*(v1f)^2 + (1/2)*m2*(v2f)^2 (equation 2)
Here, the term 0 represents the initial kinetic energy of particle 2, which is 0 since it's stationary.
Now, we can solve equations 1 and 2 simultaneously to find v1f and v2f:
m1*v1i = m1*v1f + m2*v2f (equation 1)
(1/2)*m1*(v1i)^2 = (1/2)*m1*(v1f)^2 + (1/2)*m2*(v2f)^2 (equation 2)
Substituting the given values:
0.29 kg * 1.8 m/s = 0.29 kg * v1f + 0.40 kg * v2f (equation 1)
0.5 * 0.29 kg * (1.8 m/s)^2 = 0.5 * 0.29 kg * (v1f)^2 + 0.5 * 0.40 kg * (v2f)^2 (equation 2)
After solving equations 1 and 2, we find:
v1f = -1.1 m/s (negative sign indicates particle 1 moving in the opposite direction)
v2f = 1.3 m/s
Now, we can find the time it takes for particle 2 to reach the wall using the equation:
x = vt
where x is the distance traveled, v is the velocity, and t is the time taken.
Using the given distance xw = 70 cm = 0.7 m and velocity v2f = 1.3 m/s, we can solve for t:
t = x / v
t = 0.7 m / 1.3 m/s
t ≈ 0.5385 s
Since the wall is at x = 0, we know that particle 2 will take the same amount of time to come back to the collision point, and its velocity will be in the opposite direction.
Now, we can calculate the position on the x-axis where particle 2 collides with particle 1:
Position = Initial position + (Velocity * Time)
Position = 0 + (v2f * 2t) (since it travels twice the distance)
Substituting the values:
Position = 0 + (1.3 m/s * 2 * 0.5385 s)
Position ≈ 1.402 m
Therefore, particle 2 collides with particle 1 at approximately x = 1.402 meters on the x-axis.