Question #213394

Two balls A and B of masses 10g and 20g moves on x axis towards each other with equal speed. They collide and stick together. What percentage of kinetic energy is lost?


1
Expert's answer
2021-07-05T08:42:28-0400

ma=10g=0.01kgm_a= 10g=0.01kg

mb=20g=0.02kgm_b= 20g=0.02kg

va=vb\vec v_a= -\vec v_b

according to the law of conservation of momentum:\text{according to the law of conservation of momentum:}

mava+mbvb=(ma+mb)vm_a\vec v_a+m_b\vec v_b=(m_a+m_b)\vec v

mavb+mbvb=(ma+mb)v-m_a\vec v_b+m_b\vec v_b=(m_a+m_b)\vec v

v=mbmama+mbvb\vec v=\frac{m_b-m_a}{m_a+m_b}\vec v_b

v=vb3\vec v=\frac{\vec v_b}{3}

before collision\text{before collision}

Ek=mava22+mbvb22E_k= \frac{m_av_a^2}{2}+\frac{m_bv_b^2}{2}

Ek=mavb22+mbvb22E_k= \frac{m_av_b^2}{2}+\frac{m_bv_b^2}{2}

Ek=vb2(ma+mb)2=0.015vb2E_k= \frac{v_b^2(m_a+m_b)}{2}=0.015v_b^2

after collision:\text{after collision:}

Ek=v2(ma+mb)2=vb2(ma+mb)18=0.01vb26E'_k= \frac{v^2(m_a+m_b)}{2}=\frac{v_b^2(m_a+m_b)}{18}=\frac{0.01v_b^2}{6}

p=EkEkEk=1EkEkp = \frac{E_k-E'_k}{E_k}=1-\frac{E'_k}{E_k}

p=10.01vb260.015vb20.89=89%p = 1- \frac{0.01v_b^2}{6*0.015*v_b^2}\approx0.89=89 \%

Answer: 89% of kinetic energy is lost\text{Answer: 89\% of kinetic energy is lost}


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