Question

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

Answer 1

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

(b)

$m_1v_1=(m_1+m_2)v_x\to v_x=v_1\frac{m_1}{m_1+m_2}=5\cdot\frac{90}{90+95}=2.43(m/s)$

$m_2v_2=(m_1+m_2)v_y\to v_y=v_2\frac{m_2}{m_1+m_2}=3\cdot\frac{95}{90+95}=1.54(m/s)$

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

$\alpha=\arctan(\frac{v_y}{v_x})=\arctan(\frac{1.54}{2.43})=32.3°$

(c)

$\Delta K=\frac{1}{2}(m_1+m_2)v^2-(\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2)=$

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