Two particles of mass m 1 = 1.2 kg and m 2 = 3.2 kg undergo a one-dimensional he

Question

Two particles of mass m1 = 1.2 kg and m2 = 3.2 kg undergo a one-dimensional head-on collision as shown in the figure below. Their initial velocities along x are v1i = 10 m/s and v2i = -7.4 m/s. The two particles stick together after the collision (a completely inelastic collision). (Assume to the right as the positive direction.)

Two particles of mass m1 = 1.2 kg and m2 = 3.2 kg undergo a one-dimensional head-on collision as shown in the figure below. Their initial velocities along x are v1i = 10 m/s and v2i = -7.4 m/s. The two particles stick together after the collision (a completely inelastic collision). (Assume to the right as the positive direction.) Find the velocity after the collision. How much kinetic energy is lost in the collision?

Explanation / Answer

a) by momentum balance

m1v1+m2v2= (m1+m2)*v

==> v=(1.2*10-3.2*7.4)/(1.2+3.2)=-2.6545 m/s

b) change in kinetic energy=0.5*1.2*10^2+0.5*3.2*7.4^2- 0.5*4.4*2.6545^2=132.11 J

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