You wonder what would happen if you fired a 9.109e-31kg particle moving at 20 m/s in the x direction of a 1.672e-24kg particle moving at 12 m/s at 110 degrees from the x direction. The lighter particle hits the screen at a 100 degree angle from the x direction. What is the final velocities of both particles? Assuming a classical, perfectly elastic collision.
To solve this problem, we can apply the principles of conservation of momentum and conservation of kinetic energy in an elastic collision.
Let's denote the lighter particle as particle A and the heavier particle as particle B.
Given:
Mass of particle A (ma) = 9.109e-31 kg
Initial velocity of particle A (va) = 20 m/s (in the x-direction)
Mass of particle B (mb) = 1.672e-24 kg
Initial velocity of particle B (vb) = 12 m/s (making an angle of 110 degrees with the x-direction)
Final angle made by particle A with the x-direction = 100 degrees
Since momentum is conserved in the x-direction, we have:
(ma * va) + (mb * vb) = (ma * vaf) + (mb * vbf)
(9.109e-31 kg * 20 m/s) + (1.672e-24 kg * 12 m/s) = (9.109e-31 kg * vaf) + (1.672e-24 kg * vbf) --(1)
Now, let's find the x and y components of the velocity of particle B:
vb_x = vb * cos(110 degrees) --> x-component of velocity of particle B
vb_y = vb * sin(110 degrees) --> y-component of velocity of particle B
vb_x = 12 m/s * cos(110 degrees) = -5.699 m/s (negative sign indicates velocity opposite to the x-direction)
vb_y = 12 m/s * sin(110 degrees) = 10.935 m/s (positive sign indicates velocity in the positive y-direction)
Since kinetic energy is also conserved in an elastic collision, we have:
(1/2 * ma * va^2) + (1/2 * mb * vb^2) = (1/2 * ma * vaf^2) + (1/2 * mb * vbf^2) --(2)
Substituting the x and y components of the velocities into equation (1), we have:
(9.109e-31 kg * 20 m/s) + (1.672e-24 kg * (-5.699 m/s)) = (9.109e-31 kg * vaf) + (1.672e-24 kg * vbf_x) --(3)
(9.109e-31 kg * 20 m/s) + (1.672e-24 kg * 10.935 m/s) = (9.109e-31 kg * vaf) + (1.672e-24 kg * vbf_y) --(4)
Now, using the information about the final angle made by particle A with the x-direction, we can calculate the final velocity components of particle A as follows:
vaf_x = vaf * cos(100 degrees) --> x-component of velocity of particle A
vaf_y = vaf * sin(100 degrees) --> y-component of velocity of particle A
Finally, we can solve equations (3) and (4) simultaneously to find vaf and vbf.
To determine the final velocities of both particles after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy for an elastic collision.
1. Calculate the initial momentum of each particle.
- For the first particle (9.109e-31 kg) moving at 20 m/s in the x direction, the initial momentum is: (9.109e-31 kg) * (20 m/s) = 1.8218e-29 kg m/s in the x direction.
- For the second particle (1.672e-24 kg) moving at 12 m/s at 110 degrees from the x direction, we need to find the x and y components of its momentum.
- The x-component of momentum is: (1.672e-24 kg) * (12 m/s) * cos(110 degrees) = -1.4769e-23 kg m/s.
- The y-component of momentum is: (1.672e-24 kg) * (12 m/s) * sin(110 degrees) = -3.7858e-24 kg m/s.
2. Calculate the total initial momentum of the system.
- The total initial momentum is the vector sum of the initial momenta of both particles, calculated by adding their x and y components separately.
- The x-component of the total initial momentum is: 1.8218e-29 kg m/s - 1.4769e-23 kg m/s = -1.4767e-23 kg m/s.
- The y-component of the total initial momentum is: 0 kg m/s - 3.7858e-24 kg m/s = -3.7858e-24 kg m/s.
3. Calculate the final velocities of the particles using conservation of momentum and conservation of kinetic energy.
- Since the collision is elastic, both momentum and kinetic energy are conserved.
- The final momentum of each particle can be expressed as the product of its mass and final velocity.
- We can use the equations:
- m1 * v1f_x + m2 * v2f_x = m1 * v1i_x + m2 * v2i_x (conservation of momentum in the x direction)
- m1 * v1f_y + m2 * v2f_y = m1 * v1i_y + m2 * v2i_y (conservation of momentum in the y direction)
- 0.5 * m1 * v1f^2 + 0.5 * m2 * v2f^2 = 0.5 * m1 * v1i^2 + 0.5 * m2 * v2i^2 (conservation of kinetic energy)
4. Solve the system of equations to find the final velocities.
- Substituting the given values into the equations, we can solve for v1f_x, v1f_y, v2f_x, v2f_y using algebraic manipulation.
- Once we have the components of the final velocities, we can calculate their magnitudes using pythagorean theorem.
It is important to note that the calculation involves trigonometric functions and vector components, which can make it more complex. You can use a scientific calculator or programming language to evaluate the trigonometric functions and perform the necessary calculations.