Question #152487

a car of mass 1250 kg and a bus of mass 5000 kg are moving with eqaul linear momentum. find the ratio of their kinetic energies


1
Expert's answer
2020-12-23T10:25:42-0500

By definition, the linear momentum of a body is:


p=mvp = mv

where mm is the mass of a body, and vv is its velocity. Since both cars have equal linear momentum, we can write:


p1=p2m1v1=m2v2p_1 = p_2\\ m_1v_1 = m_2v_2

where m1=1250kg,m2=5000kgm_1 = 1250kg, m_2 = 5000kg are the masses, and v1,v2v_1, v_2 are the velocities of the first and second cars respectively. Thus, obtain the ratio:


v1v2=m2m1 v12v22=m22m12\dfrac{v_1}{v_2} = \dfrac{m_2}{m_1}\\ \space \\ \dfrac{v_1^2}{v_2^2} = \dfrac{m_2^2}{m_1^2}

On the other hand, their kinetic energies are given as follows:


K1=m1v122K2=m2v222K_1 = \dfrac{m_1v_1^2}{2}\\ K_2 = \dfrac{m_2v_2^2}{2}

Their ratio is:


K1K2=m1v12m2v22\dfrac{K_1}{K_2} = \dfrac{m_1v_1^2}{m_2v_2^2}

Substituting the expression for v12/v22v_1^2/v_2^2, obtain:


K1K2=m1m22m2m12=m2m1K1K2=5000kg1250kg=4\dfrac{K_1}{K_2} = \dfrac{m_1\cdot m_2^2}{m_2\cdot m_1^2} = \dfrac{m_2}{m_1}\\ \dfrac{K_1}{K_2} =\dfrac{5000kg}{1250kg} = 4

Answer. 4.


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