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

a)
`V_A = ((M_A - eM_B)U_A + M_BU_B(1+e))/(M_A + M_B)`

`V_B = (M_AU_A(1+e) + (M_B - eM_A)U_B)/(M_A + M_B)`

b)
`U_A = 3.66 m/s`

`V_B = 4.32 m/s`

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Step-by-step explanation:

Let
`U_A` be the initial velocity of rod A

Let
`U_B` be the initial velocity of rod B

Let
`V_A` be the final velocity of rod A

Let
`V_B` be the final velocity of rod B

Using the principle of conservation of momentum:

`M_AU_A + M_BU_B = M_AV_A + M_BV_B`............(1)

Coefficient of restitution,
`e = (V_B - V_A)/(U_A - U_B)`

`V_A = V_B - e(U_A - U_B)`........................(2)

Substitute equation (2) into equation (1)

`M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B`..............(3)

Solving for
`V_B` in equation (3) above:

`V_B = (M_AU_A(1+e) + (M_B - eM_A)U_B)/(M_A + M_B)`....................(4)

From equation (2):

`V_B = V_A + e(U_A -U_B)`......(5)

Substitute equation (5) into (1)

`M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))`..........(6)

Solving for
`V_A` in equation (6) above:

`V_A = ((M_A - eM_B)U_A + M_BU_B(1+e))/(M_A + M_B)`.........(7)

b)

`M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?`

Rod A is said to be at rest after the impact,
`V_A = 0 m/s`

Substitute these parameters into equation (7)

`0 = ((2 - 0.65*1)U_A - (1*3)(1+0.65))/(2+1)\\U_A = 3.66 m/s`

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

`V_B = ((2*3.66)(1+0.65) - (1 - (0.65*2))*3)/(2+1)\\V_B = 4.32 m/s`

c) Impulse,
`I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)`

`I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))`

I = 0
`kg m/s^2`

d) %
`\triangle KE = ((0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2))/(0.5 M_A U_A^2 + 0.5 M_B U_B^2) * 100\%`

%
`\triangle KE = (((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2)))/( (0.5 *2*3.66^2) + 0.5*1*(-3)^2)) * 100\%`

%
`\triangle KE = -47.85 \%`