Let\'s put some numbers to the previous problem. Let M_1 = 5kg and M_2 = 10kg wi

Question

Let's put some numbers to the previous problem. Let M_1 = 5kg and M_2 = 10kg with velocities upsilon_1 = 10m/s and upsilon_2 = - 2m/s. They collide elastically and exit the collision with velocities upsilon'_1 and upsilon'_2. (a)What is the total momentum of the system? (b)What is the total mechanical energy of the system? (c)Using conservation of momentum and conservation of mechanical energy, what are upsilon'_1 and upsilon'_2? (d)Alternatively, consider the case where the collision is completely inelastic, that is, when the two objects move together as one after the collision. What is the velocity of the M_1 - M_2 combined system? (e)How much kinetic energy is lost in the completely inelastic situation?

Explanation / Answer

a)

Total initial Momentum is given as

Pi = M1 v1 + M2 v2 = 5 (10) + (10) (-2) = 30 kgm/s

b)

Total initial mechanical energy is given as

KEi = (0.5) (M1 v21 + M2 v22 ) = (0.5) (5 (10)2 + (10) (-2)2) = 270 J

c)

using conservation of energy

M1 v1 + M2 v2 = M1 v'1 + M2 v'2

30 = 5 v'1 + 10 v'2

v'1 = (30 - 10 v'2 )/5 eq-1

using conservation of mechanical energy

(0.5) (M1 v21 + M2 v22 ) = (0.5) (M1 v'21 + M2 v'22 )

270 = (0.5) (5 ((30 - 10 v'2 )/5)2 + 10 v'22 )

v2' = 6 m/s

using eq-1

v'1 = (30 - 10 v'2 )/5 = (30 - 10 (6) )/5 = - 6 m/s

d)

let assume the final velocity of the combination after collision as "V"

using conservation of momentum when collision is inelastic

M1 v1 + M2 v2 = (M1 + M2) V

30 = (5 + 10) V

V = 2 m/s

e)

Kinetic energy lost = (0.5) (M1 v21 + M2 v22 ) - (0.5) (M1 + M2 ) V2 = 270 - (0.5) (15) (2)2 = 240 J

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