. Two rods, with masses MA and MB having a coefficient of restitution, e, move along a common line on a surface, figure 2.
a) Find the general expression for the velocities of the two rods after impact.
(7 marks)
b) If mA = 2 kg, m3 = 1 kg, Vo= 3 m/s, and e = 0.65, find the value of the initial
velocity v4 of rod A for it to be at rest after the impact and the final velocity vB
of rod B.
(4 marks)
c) Find the magnitude of the impulse for the condition in part b. (3 marks)
d) Find the percent decrease in kinetic energy
which corresponds to the impact in
(3 marks)
part b.

a)

b)

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Explanation:

Let be the initial velocity of rod A

Let be the initial velocity of rod B

Let be the final velocity of rod A

Let be the final velocity of rod B

Using the principle of conservation of momentum:

............(1)

Coefficient of restitution,

........................(2)

Substitute equation (2) into equation (1)

..............(3)

Solving for in equation (3) above:

....................(4)

From equation (2):

......(5)

Substitute equation (5) into (1)

..........(6)

Solving for in equation (6) above:

.........(7)

b)

Rod A is said to be at rest after the impact,

Substitute these parameters into equation (7)

To calculate the final velocity, , substitute the given parameters into (4):

c) Impulse,

I = 0

d) %

%

%

answer:

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explanation: