Question

Two particles of mass m1 = 2.0 kg and m2 = 2.6 kg undergo a one-dimensional head-on collision as shown in the figure below. Their initial velocities along x are vii = 15 m/s and v2,--6.8 m/s. The two particles stick together after the collision (a completely inelastic collision. (Assume to the right as the positive direction.) mi m2 (a) Find the velocity after the collision. 2.6782 m/s (b) How much kinetic energy is lost in the collision? 153.907x

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Answer #1

I think the combined velocity of the two particles is determined by applying conservation of momentum by you. And I am considering the value of final velocity, v = 2.6782 m/s is correct.

Now for part (b) -

Total initial energy, KEi = kinetic energy of first particle + kinetic energy of the second particle

= (1/2)*m1*v1^2 + (1/2)*m2*v2^2

= 0.5*2.0*15^2 + 0.5*2.6*(-6.8)^2

= 225 + 60.11 = 285.11 J

Total final energy of the two particle after collision, KEf = (1/2)*(m1+m2)*v^2

= 0.5*(2.0+2.6)*2.6782^2

= 16.50 J

Therefore, amount of kinetic energy lost in the collision = KEi - KEf

= 285.11 J - 16.50 J = 268.61 J (Answer)

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