Question #142698

A 3-kg object moving to the right on a frictionless, horizontal surface with a speed of 2 m/s collides head on and sticks to a 2-kg object that is initially moving to the left with a speed of 4 m/s. After the collision, which statenoent is true?
(a) The kinetic energy of the system is 20 J. (b) The momentum of the system is 14 kgm/s. (c)
The kinetic energy of the system is greater than 5 J but less than 20 J. (d) The momentum of the system is 22?

Expert's answer

The kinetic energy of the system is (assuming that all kinetic energy of both objects after collision converts into kinetic energy of the system):


E=mv2/2=m1v12/2+m2v22/2,E=mv^2/2=m_1v_1^2/2+m_2v_2^2/2, (1)


where m - the mass of the system, that equal m1+m2m_1+m_2 ;

vv - the speed of the system;

m1m_1 - the mass of the first object;

v1v_1 - the speed of the first object;

m2m_2 - the mass of the second object;

v2v_2 - the speed of the second object.


E=3⋅22/2+2⋅42/2=22 J.E=3\cdot 2^2/2+2\cdot 4^2/2=22\space J.


The momentum of the system is 


P=mv=(m1+m2)v.P=mv=(m_1+m_2)v.


From (1)


v=m1v12+m2v22m1+m2=3⋅22+2⋅423+2=2.97 m/s.v=\sqrt {\frac {m_1v_1^2+m_2v_2^2}{m_1+m_2}}=\sqrt {\frac {3\cdot 2^2+2\cdot 4^2}{3+2}}=2.97\space m/s.


P=(2+3)⋅2.97=14.83 kg⋅m/s.P=(2+3)\cdot 2.97=14.83\space kg\cdot m/s.

Answer: (b) The momentum of the system is 14 kgm/s (approximately!!!).


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