Question

Determine the velocities of A and B after impact. The speed of A and B prior to impact 10 m/s and 5 m/s, respectively. The direction of each velocity prior to impact is given in the figure below. The mass of both particles is 1 kg. The coefficient of restitution is 0.75. 13 12 5 ch 3 4 A B

Answer 1

v'. mi em2 та + т. V1 + (1+e)т. та + т. V2 v'. т2 еті та + т. V2 + (1 + еті та + m2 V1

This is the formula for velocities after inelastic collision
m1 = m2 = 1kg
e = 0.75

13 12 5 3 5 у 4 А.

For horizontal velocity:
cos(x) = 4/5
cos(y) = 5/13

V_a = [(1-0.75)/2]10cos(x) + [(1+0.75)/2](-5cos(y))
V_a = -0.68 m/s

Vb = [(1-0.75)/2](-5cos(y)) + [(1+0.75)/2](10cos(x))
Vb = 6.76 m/s

For vertical velocity:
sin(x) = 3/5
sin(y) = 12/13

V_a = [(1-0.75)/2]10sin(x) + [(1+0.75)/2](-5sin(y))
V_a = -16.75 m/s

Vb = [(1-0.75)/2](-5sin(y)) + [(1+0.75)/2](10sin(x))
Vb = 4.67 m/s

Final velocities (magnitude):
A - [(-0.68)² + (-16.75)²]^1/2 = 16.76 m/s

B - [(6.76)² + (4.67)²]^1/2 = 8.22 m/s