Question #289984

In a road accident, a van of mass 1500 kg moving at a speed of 28m * s ^ - 1 ran into the back of a car of mass 900 kg moving in the same direction at a speed of 11m * s ^ - 1 . As a result of the impact, the car was pushed forward at a speed of 18m * s ^ - 1 .




a Calculate the velocity of the van immediately after the impact.




b Calculate:




i the loss of kinetic energy of the van, ii the gain of kinetic energy of the car,




iii the total change of kinetic energy of the two vehicles.

Expert's answer

a. By conservation of momentum:


mv+MV=mu+MU, U=mv+MVmuM=23.4 m/s.mv+MV=mu+MU,\\\space\\ U=\dfrac{mv+MV-mu}{M}=23.4\text{ m/s}.


b i. The loss of kinetic energy of the van:


ΔKEv=12M(U2V2)=177330 J.\Delta KE_v=\dfrac12M(U^2-V^2)=177330\text{ J}.

b. ii. The gain of kinetic energy of the car:


ΔKEc=12m(u2u2)=91350 J.\Delta KE_c=\dfrac12m(u^2-u^2)=91350\text{ J}.

b. iii. The total change of kinetic energy of the two vehicles:


ΔKEc=ΔKEv+ΔKEc=268680 J.\Delta KE_c=\Delta KE_v+\Delta KE_c=-268680\text{ J}.



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