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?
How do i set this problem up?
The first collision (conservation of energy, momentum) gives you the speeds of M1 and m2 after the collision. Then, the distance m2 travels is 70 cm + (70-x) cm. But it does it in the same time as m1 travel x cm. You have the distances, velocitys, and times, solve for x.
ok so velocity of m1 = 150 m/s and m2 = -29 m/s.....where do i go from there?
To set up this problem, we will use the principles of conservation of momentum and conservation of kinetic energy. We can break down the problem into three stages:
Stage 1: The initial state
Particle 1 of mass m1 is moving rightward along the x-axis with a speed of 1.8 m/s. We can denote its velocity as v1i = 1.8 m/s, and its initial position as x1i = 0 m. Particle 2 is stationary, so its velocity is v2i = 0 m/s.
Stage 2: The collision
When particle 1 reaches x = 0, it undergoes an elastic collision with particle 2. In an elastic collision, both momentum and kinetic energy are conserved. Let's denote the final velocities of particle 1 and 2 after the collision as v1f and v2f, respectively.
Stage 3: Particle 2's motion
After bouncing off the wall at xw = 70 cm (or 0.7 m), particle 2 continues moving towards the right. Since there is no loss of speed, its final velocity will be the same as its initial velocity after the collision. We can denote this as v2f = v2f'.
To find the position on the x-axis where the two particles collide again, we need to determine the final positions of both particles.
Now, let's go step by step to solve the problem.