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## 73. Two objects, A and B, with masses of 3.2 kg and 1.8 kg, move on a frictionless horizontal surface. Object A moves to the right at a cons

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73. Two objects, A and B, with masses of 3.2 kg and 1.8 kg, move on a frictionless horizontal surface. Object A moves to the right at a constant speed of 5.1 m/s while object B moves to the right at a constant speed 1.4 m/s. They collide and stick together (a perfectly inelastic collision).

a. Determine the total momentum of the system (both objects) before the collision

b. Determine the total kinetic energy of the system before the collision

c. *Find the speed of the two objects after the collision

d. *Find the total kinetic energy of the system after the collision.

e. *Is the kinetic energy of the system conserved? Explain.

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Physics
3 years
2021-08-15T04:52:40+00:00
2021-08-15T04:52:40+00:00 1 Answers
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## Answers ( )

(a) The

total momentum of the systembefore the collision is 18.84 kg m/s.(b)

The total kinetic energyof the systembefore collision is 43.384 J.(c) The

speed of the objectaftercollisionwill be3.77 m/s.(d) The

total kinetic energyof the systemafter collisionis35.5 J.(e) The

kinetic energyof the system isnot conserved.Explanation:When two objects collide or hit each other, they may undergo

elastic collision or inelastic collision. Both of these collision will obey thelaw of conservation of momentum.That is thesum of momentum of the objects before collisionwill be equal to thesum of momentum of the objects after collision. But there will beloss of kinetic energyin case ofinelastic collision, while there will beno loss of kinetic energy forelastic collision.Thus, in aperfect inelastic collision, both theobjects will stick together after the collision.So, here the mass of object A (Mₐ) is given as 3.2 kg and the mass of object B (Mb) is given as 1.8 kg, the speed of object A is 5.1 m/s and the speed of object B is given as 1.4 m/s. Both the objects are travelling in same direction.

(a)

Momentum of object A = Mass × Velocity = 3.2 × 5.1 =16.32 kgm/sMomentum of object B = 1.8×1.4=2.52 kgm/s.Thus, the

total momentum of the systembefore the collision is 18.84 kg m/s.(b)

Total kinetic energy of the systembefore collision can be obtained by thesum of kinetic energy of object A and kinetic energy of object B.So,

the total kinetic energy of the system = 41.62 + 1.764 =43.384 J.Thus,

the total kinetic energyof the systembefore collision is 43.384 J.(c) As after collision both the object got stick together, so the

speed of two objectswill be considered as thespeed of one objector theresultant speed after collision.So in general, theresultant speedafter collision is obtained as theratio of total momentum before collision to the sum of the masses.So, the

speed of the objectaftercollisionwill be3.77 m/s.(d)

As the sum of the masses of both the objects is 5 kg and the resultant velocity or speed is given as V = 3.77 m/s, then

Thus, the

total kinetic energyof the systemafter collisionis35.5 J.(e) So, it can be seen that the kinetic energy of the system is not conserved as the

kinetic energyof the systembefore collisionwas43.384 Jand thekinetic energy after collisionis found to be35.5 J.So there is a loss in the kinetic energy. Thus, thekinetic energyof the system isnot conserved.