Question

A 75.0-kg fullback running east with a speed of 4.80 m/s is tackled by a 97.0-kg opponent running north with a speed of 3.00 m/s. (a) Why does the tackle constitute a perfectly inelastic collision? This answer has not been graded yet. (b) Calculate the velocity of the players immediately after the tackle. magnitude direction m/s o north of east (c) Determine the mechanical energy that is lost as a result of the collision. (d) Where did the lost energy go?

Answer 1

Over a short time interval of the collision: external forces have no time to import significant impulse to the players. The two players move together after the tackle, so the collision is completely inelastic. P_xf = Sigma P_xi = > (m_1 + m_2) v_f cos theta = m_1 v_1i + 0 v_f cos theta = 75 (4.8)/75 + 97 = 2.093 m/s P_yf = Sigma P_yi = > (m_1 + m_2) v_f sin theta = 0 + m_2 v_2i v_f sin theta = 97 (3)/75 + 97 = 1.691 m/s v_f^2 (sin^2 theta + cos^2 theta) = v_f^2 = (1.691)^2 + (2.093)^2 v_f = 2.69 m/s tan theta = v_f sin theta/v_f cos theta = 1.691/2.093 = 0.807931199 theta = 38.93 degree north of east KE_lost = 1/2 m_1 v_1i^2 + 1/2 m_2 v_2i^2 - 1/2 (m_1 + m_2) v_f^2 = 1/2 [(75) (4.8)^2 + 97 (3)^2] - 1/2 (75 + 97) (2.69)^2 = 678.19 J