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!
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