Question

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. 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. 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 part b.​

Answer 1

Answer:

A.) Find the answer in the explanation

B.) Ua = 7.33 m/s , Vb = 7.73 m/s

C.) Impulse = 17.6 Ns

D.) 49%

Explanation:

Let Ua = initial velocity of the rod A

Ub = initial velocity of the rod B

Va = final velocity of the rod A

Vb = final velocity of the rod B

Ma = mass of rod A

Mb = mass of rod B

Given that

Ma = 2kg

Mb = 1kg

Ub = 3 m/s

Va = 0

e = restitution coefficient = 0.65

The general expression for the velocities of the two rods after impact will be achieved by considering the conservation of linear momentum.

Please find the attached files for the solution