Question #90737
A 1.3-ton sedan car travellng west at 72 km/h runs into a 2.5-ton SUV travelling south at 36 km/h at an intersection. Given that the cars stick together after the collision, their speed and the direction are:
1
Expert's answer
2019-06-17T12:54:18-0400

The law of conservation of momentum gives

m1v1=(m1+m2)vcosθm_1v_1=(m_1+m_2)v \cos\theta

m2v2=(m1+m2)vsinθm_2v_2=(m_1+m_2)v \sin\theta

So

tanθ=m2v2m1v1=2500kg×10m/s1300kg×20m/s=0.9615\tan\theta=\frac{m_2v_2}{m_1v_1}=\frac{2500\:\rm{kg}\times 10\:\rm{m/s}}{1300\:\rm{kg}\times 20\:\rm{m/s}}=0.9615

θ=43.9(southofwest)\theta=43.9^{\circ}\quad\rm{(south\: of\: west)}

v=1m1+m2(m1v1)2+(m2v2)2v=\frac{1}{m_1+m_2}\sqrt{(m_1v_1)^2+(m_2v_2)^2}

=11300kg+2500kg(1300kg×20m/s)2+(2500kg×10m/s)2=\frac{1}{1300\:\rm{kg}+2500\:\rm{kg}}\sqrt{(1300\:\rm{kg}\times 20\:\rm{m/s})^2+(2500\:\rm{kg}\times 10\:\rm{m/s})^2}

=9.5m/s=34km/h=9.5\:\rm{m/s}=34\:\rm{km/h}


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