Bouncing Particles on a smooth table

[Bucky]

"Particles A, B and C , each of mass m, lie at rest in a straight line in the order stated. A is projected directly towards B with velocity u. The coefficient of restitution is 0.5 in each impact that follows. Show that there will be three impacts in total and find the final velocities of each particle."

I have 'an' answer for the first part:-

A and B Collide

Velocity of separation [tex]= e (u_1 - u_2)
=0.5(u-0)
=0.5u
[/tex]
[tex]
v_1 + v_2 = u //
v_1 = u - v_2
[/tex]

[tex]
v_2 - v_1 = e(u_1 - u_2)//
v_2 - u + v_2 = 0.5u//
2v_2 = 1.5u//
v_2 = 0.75u
[/tex]
[tex]
v_1 = u - v_2
v_1 = u - 0.75u
v_1 = 0.25u
[/tex]

is this the right way to do this question? the problem comes later when the moving particles hit each other.

 

[anathema]

Hello! Thank you for your contribution to the forum post. Your approach to solving the problem is correct so far. Now, let's move on to the next step where the moving particles collide with each other.

After the first impact, particle A will have a final velocity of 0.25u and particle B will have a final velocity of 0.75u. Now, particle B will continue moving towards particle C with a velocity of 0.75u. When it collides with particle C, the same process will occur again.

Particle B and C Collide

Velocity of separation = 0.5(0.75u - 0)
= 0.375u

v_2 + v_3 = 0.75u
v_2 = 0.75u - v_3

v_3 - v_2 = 0.375u
v_3 - 0.75u + v_3 = 0.375u
2v_3 = 0.375u + 0.75u
v_3 = 0.5625u

v_2 = 0.75u - v_3
v_2 = 0.75u - 0.5625u
v_2 = 0.1875u

Therefore, after the second impact, particle B will have a final velocity of 0.1875u and particle C will have a final velocity of 0.5625u. Now, particle C will continue moving with a velocity of 0.5625u towards particle A.

Particle C and A Collide

Velocity of separation = 0.5(0.5625u - 0.25u)
= 0.15625u

v_3 + v_1 = 0.5625u
v_3 = 0.5625u - v_1

v_1 - v_3 = 0.15625u
v_1 - 0.5625u + v_3 = 0.15625u
2v_1 = 0.15625u + 0.5625u
v_1 = 0.359375u

v_3 = 0.5625u - v_1
v_3 = 0.5625u - 0.359375u
v_

 

[wais]

Yes, your approach to finding the velocities after the first collision between A and B is correct. However, there are a few things to consider for the subsequent collisions.

First, after the first collision, particle B will be moving with a velocity of 0.75u and particle A will be moving with a velocity of 0.25u. This means that when A reaches B for the second collision, their velocities will be different from the initial conditions given in the problem.

Second, since the particles are on a smooth table, there will be no external forces acting on them to change their velocities between collisions. This means that the velocities of particles A and B will remain constant throughout the collisions.

Therefore, for the second collision, the velocities of particles A and B will be 0.25u and 0.75u respectively. Using the same approach as before, we can find the velocities after the second collision to be 0.125u and 0.625u for particles A and B respectively.

For the third collision, the velocities of particles A and B will be 0.125u and 0.625u respectively. Using the same approach again, we can find the final velocities to be 0.0625u and 0.5625u for particles A and B respectively.

Since particle C is not involved in any of the collisions, its final velocity will remain 0.

Therefore, there will be three impacts in total and the final velocities of particles A, B, and C will be 0.0625u, 0.5625u, and 0 respectively.