Answer to Question #172925 in Physics for Margarette

Question #172925

A car with a mass of 2500 kg and moving at 3 m/s northeast collides with a 3000 kg car moving northwest at 2.5 m/s . Find their final velocities. Assume a perfectly inelastic collision.


1
Expert's answer
2021-03-23T08:15:05-0400

Let's choose the axes as shown in the figure below.



In a perfectly inelastic collision the momentum conservation law still holds. Thus, the total momentum before the collision in the projection on the x-axis is:


px=m1v1p_{x} = -m_1v_1

is equal to the total momentum after the collision in the projection on the x-axis:


px=(m1+m2)vxp_{x}' = (m_1 + m_2)v_x'

where m1=2500kg,m2=3000kgm_1 = 2500kg, m_2 = 3000kg are the masses of the cars, v1=3m/s,v2=2.5m/sv_1 = 3m/s, v_2 = 2.5m/s are their speeds before the collision, and vxv_x' is their speed after the collsion in x-direction. Thus, obtain:


m1v1=(m1+m2)vxvx=m1v1m1+m2=250032500+30001.36m/s-m_1v_1 = (m_1 + m_2)v_x'\\ v_x' = -\dfrac{m_1v_1}{m_1 + m_2} = -\dfrac{2500\cdot 3}{2500+3000} \approx -1.36m/s

Similarly, the momentum conservation law in the projection on y-axis gives:


m2v2=(m1+m2)vym_2v_2 = (m_1 + m_2)v_y'

where vyv_y' is car's speed after the collsion in y-direction. Thus, obtain:


m2v2=(m1+m2)vyvy=m2v2m1+m2=30002.52500+30001.36m/sm_2v_2 = (m_1 + m_2)v_y'\\ v_y' = -\dfrac{m_2v_2}{m_1 + m_2} = -\dfrac{3000\cdot 2.5}{2500+3000} \approx 1.36m/s

Thus, the final velocity is:


v=(1.36,1.36) m/s\mathbf{v}' = (-1.36, 1.36)\space m/s

Answer. v=(1.36,1.36) m/s\mathbf{v}' = (-1.36, 1.36)\space m/s.


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