Partv1-v2cle A and B have masses 0.75 and 0.65 respectively. They are approaching each other moving along a straight line with speed 6m/s and 4m/s respectively. The coefficient between them is 4/3. After collision,the direction of motion of B is reversed. Particle B stikes a fixed vertical wall at right angle. If the coefficient of restitution between B and the wall is 2/3. Show that B will colide again with a and find the speed of A and B after the second collision.

Solution
M1U1 + M2U2=M1V1 + M2V2
0.75×6+0.65×-4=0.75V1 + 0.65V2
1.9=0.75v1+0.65v2............(1)

e=v1-v2/u1-u2
4/3=v1-v2/6--4
4/3=v1-v2/10
10×4/3=v1-v2
40/3=v1-v2..... (2)
Solving (1) and (2) simultaneously
V1 = 7.53 and V= -5.77
How can i solve the question with this information!
Please help me

To solve the question, you need to find the values of V1 and V2, which are the velocities of particles A and B after the collision, and then determine if B will collide with A again.

From the given information, you have:

M1 = 0.75 kg (mass of particle A)
M2 = 0.65 kg (mass of particle B)
U1 = 6 m/s (initial velocity of A)
U2 = 4 m/s (initial velocity of B)
e = 4/3 (coefficient of restitution between A and B)

Using the equation: M1U1 + M2U2 = M1V1 + M2V2, you can calculate V1 and V2:

0.75 × 6 + 0.65 × (-4) = 0.75V1 + 0.65V2
4.5 - 2.6 = 0.75V1 + 0.65V2
1.9 = 0.75V1 + 0.65V2 ... (1)

Next, using the equation: e = (V1 - V2)/(U1 - U2), you can calculate V1 - V2:

4/3 = (V1 - V2)/(6 - (-4))
4/3 = (V1 - V2)/(6 + 4)
4/3 = (V1 - V2)/10
10 × (4/3) = V1 - V2
40/3 = V1 - V2 ... (2)

Now, you can solve equations (1) and (2) simultaneously to find the values of V1 and V2:

Equation (1): 1.9 = 0.75V1 + 0.65V2
Equation (2): 40/3 = V1 - V2

Solving these equations will provide the values of V1 and V2:

V1 = 7.53 m/s
V2 = -5.77 m/s

To determine if B will collide with A again, you need to observe the direction of motion of B after the collision. Since B strikes a fixed vertical wall at a right angle and its direction of motion is reversed, B will start moving in the opposite direction. This means that B will eventually collide with A again.

To find the speed of A and B after the second collision, you can use the same initial velocities and masses, and the coefficient of restitution between B and the wall.

Using the same method as before, but with the coefficient of restitution between B and the wall (e2 = 2/3), you can calculate the velocities after the second collision.

Note that the mass of the wall is usually considered to be much larger than the mass of the particle, so it can be neglected in the calculations.

I hope this explanation helps you solve the question.