Answer to Question #135262 in Mechanics | Relativity for Ateufack Zeudom

Question #135262
A 90.0 kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0 kg opponent running north with a speed of 3.00 m/s. (a) Explain why the successful tackle constitutes a perfectly inelastic collision. (b) Calculate the velocity of the players immediately after the tackle. (c) Determine the mechanical energy that disappears as a result of the collision. Account for the missing energy
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Expert's answer
2020-10-05T10:58:51-0400

(a) Because they are stuck together, it is a perfectly inelastic collision.


(b)


m1v1=(m1+m2)vxvx=v1m1m1+m2=59090+95=2.43(m/s)m_1v_1=(m_1+m_2)v_x\to v_x=v_1\frac{m_1}{m_1+m_2}=5\cdot\frac{90}{90+95}=2.43(m/s)


m2v2=(m1+m2)vyvy=v2m2m1+m2=39590+95=1.54(m/s)m_2v_2=(m_1+m_2)v_y\to v_y=v_2\frac{m_2}{m_1+m_2}=3\cdot\frac{95}{90+95}=1.54(m/s)


v=1.542+2.432=2.88(m/s)v=\sqrt{1.54^2+2.43^2}=2.88(m/s)


α=arctan(vyvx)=arctan(1.542.43)=32.3°\alpha=\arctan(\frac{v_y}{v_x})=\arctan(\frac{1.54}{2.43})=32.3°


(c)


ΔK=12(m1+m2)v2(12m1v12+12m2v22)=\Delta K=\frac{1}{2}(m_1+m_2)v^2-(\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2)=


=12(90+95)2.882(129052+129532)=785(J)=\frac{1}{2}\cdot(90+95)\cdot2.88^2-(\frac{1}{2}\cdot90\cdot 5^2+\frac{1}{2}\cdot 95\cdot 3^2)=-785(J)


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