Question #343202

Blocks A and B are moving toward each other. A has a mass of 6.1 kg and a velocity of 1.5m/s, while B has a mass of 1.7kg and a velocity of −9.8m/s. They suffer a completely inelastic collision. During the collision, how much kinetic energy is lost?

1
Expert's answer
2022-05-20T13:19:41-0400

The law of conservation of momentum says

mAvA+mBvB=(mA+mB)um_Av_A+m_Bv_B=(m_A+m_B)u

The final velocity of system

u=mAvA+mBvBmA+mB=6.11.5+1.7(9.8)6.1+1.7=0.96m/su=\frac{m_Av_A+m_Bv_B}{m_A+m_B}\\=\frac{6.1*1.5+1.7*(-9.8)}{6.1+1.7}=-0.96\:\rm m/s

The loss in kinetic energy

ΔE=mAvA22+mBvB22(mA+mB)u22\Delta E=\frac{m_Av_A^2}{2}+\frac{m_Bv_B^2}{2}-\frac{(m_A+m_B)u^2}{2}

ΔE=6.11.522+1.79.822(6.1+1.7)0.9622\Delta E=\frac{6.1*1.5^2}{2}+\frac{1.7*9.8^2}{2}-\frac{(6.1+1.7)*0.96^2}{2}

ΔE=85J\Delta E=85\:\rm J


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