## 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

Question

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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3 years 2021-08-15T04:52:40+00:00 1 Answers 171 views 0

1. (a) The total momentum of the system before the collision is 18.84 kg m/s.

(b) The total kinetic energy of the system before collision is 43.384 J.

(c) The speed of the object after collision will be 3.77 m/s.

(d) The total kinetic energy of the system after collision is 35.5 J.

(e)  The kinetic energy of the system is not conserved.

Explanation:

When two objects collide or hit each other, they may undergo elastic collision or inelastic collision. Both of these collision will obey the law of conservation of momentum. That is the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision. But there will be loss of kinetic energy in case of inelastic collision, while there will be no loss of kinetic energy for elastic collision. Thus, in a perfect inelastic collision, both the objects 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/s

Momentum of object B = 1.8×1.4=2.52 kgm/s.

Thus, the total momentum of the system before the collision is 18.84 kg m/s.

(b) Total kinetic energy of the system before collision can be obtained by the sum 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 energy of the system before collision is 43.384 J.

(c) As after collision both the object got stick together, so the speed of two objects will be considered as the speed of one object or the resultant speed after collision. So in general, the resultant speed after collision is obtained as the ratio of total momentum before collision to the sum of the masses.

So, the speed of the object after collision will be 3.77 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 energy of the system after collision is 35.5 J.

(e) So, it can be seen that the kinetic energy of the system is not conserved as the kinetic energy of the system before collision was 43.384 J and the kinetic energy after collision is found to be 35.5 J. So there is a loss in the kinetic energy. Thus, the kinetic energy of the system is not conserved.