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 solve this problem, we need to use principles of conservation of momentum and conservation of kinetic energy.
Given:
Mass of particle 1 (m1) = 0.29 kg
Mass of particle 2 (m2) = 0.40 kg
Initial velocity of particle 1 (v1) = 1.8 m/s
Final position of particle 1 (x1f) = 0 m
Final position of particle 2 (x2f) = ? (to be determined)
Position of the wall (xw) = 70 cm = 0.7 m
Step 1: Calculate the initial momentum of particle 1.
Initial momentum is given by p1 = m1v1.
p1 = m1 * v1
= 0.29 kg * 1.8 m/s
= 0.522 kg·m/s
Step 2: Calculate the recoil velocity of particle 1 after the collision.
Since the collision is one-dimensional and elastic, the total momentum before and after the collision must be the same.
Total momentum before the collision = Total momentum after the collision
(particle 1 momentum) + (particle 2 momentum) = (particle 1 momentum after collision) + (particle 2 momentum after collision)
m1 * v1 + m2 * 0 = m1 * v1' + m2 * v2'
0.29 kg * 1.8 m/s + 0.40 kg * 0 = 0.29 kg * v1' + 0.40 kg * v2'
0.522 kg·m/s = 0.29 kg * v1' + 0.40 kg * v2'
Step 3: Apply conservation of kinetic energy to find the speed of particle 2 after the collision.
Since the collision is elastic, the total kinetic energy before and after the collision must be the same.
Total kinetic energy before the collision = Total kinetic energy after the collision
(0.5 * m1 * v1^2) + 0 = (0.5 * m1 * v1'^2) + (0.5 * m2 * v2'^2)
(0.5 * 0.29 kg * (1.8 m/s)^2) + 0 = (0.5 * 0.29 kg * (v1')^2) + (0.5 * 0.40 kg * (v2')^2)
0.4674 J = 0.5 * 0.29 kg * (v1')^2 + 0.5 * 0.40 kg * (v2')^2
Step 4: Use the equations obtained from step 2 and step 3 to solve for v1' and v2'.
From step 2: 0.522 kg·m/s = 0.29 kg * v1' + 0.40 kg * v2'
From step 3: 0.4674 J = 0.5 * 0.29 kg * (v1')^2 + 0.5 * 0.40 kg * (v2')^2
Solving these two equations will give us the values of v1' and v2'.
Step 5: Once we have the values of v1' and v2', we can calculate the position (x2f) where particle 2 collides with particle 1.
The time taken for particle 2 to reach the wall is given by:
Time = Distance / Velocity = (xw - x2f) / v2'
The time taken for particle 1 to reach the same position x2f is given by:
Time = Distance / Velocity = x2f / v1'
Since the times taken by the particles are equal (collision happening at the same time), we can equate the two time expressions:
(xw - x2f) / v2' = x2f / v1'
Solving this equation will give us the position x2f where particle 2 collides with particle 1.
Follow these steps to solve the problem and get the final answer.