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75g 600 m/s 50 kg Figure 1: Problem 1 1. A 75-g projectile traveling at 600 m/s strikes and becomes embedded in the 50-kg block, which is initially stationary. Compute the energy lost during the impact. Express your answer as an absolute value | ∆E | and as a percentage n of the original system energy E. Solution. Since the force of impact is internal to the system composed of the projectile and block and since there are no other external forces acting on the system in the horizontal line of motion, it follows that the linear momentum of the system is conserved. m1 v1 i = (m1 + m2 )vi (1) where m1 is the mass of the projectile, m2 is the mass of the block, and v is the velocity after the impact. Thus 75(10−3 )(600) = 75(10−3) + 50 v 45 v = = 0.900 m/s (2) 50.075 The original energy E is E= 1 (75)(10−3 )(6002 ) = 13500 J 2 (3) The energy after the impact is E′ = 1 75(10−3 ) + 50 (0.9002) = 20 J 2 (4) The energy lost is ∆E = E − E ′ = 13480 J n is n= ∆E 13480 = (100) = 99.9% E 13500 (5) (6) 2. Determine the magnitude HO of the angular momentum of the 2-kg sphere about point O (a) by using the vector definition of angular momentum and (b) by using an equivalent scalar approach. The center of the sphere lies in the x − y plane. Solution. (a) The position vector from O to the sphere is r = 12i + 5j m 1 (7) y 7 m/s 2 kg 45 5m 12 m x O Figure 2: Problem 2 The velocity of it is 7 cos 45◦ i + 7 sin 45◦ j 4.95i + 4.95j m/s = = v (8) The angular momentum is HO = r × mv = (12i + 5j) × 2(4.95i + 4.95j) = 69.3k kgm2 /s (b) The distance between the sphere and O is p r = 122 + 52 = 13 m (9) (10) The angle between the velocity vector and the line from O to the sphere is 5 −1 = 22.62◦ (11) θ = tan 12 The equivalent scalar of angular momentum then is rmv sin (45◦ − 22.62◦) = 69.3 kgm2 /s (12) 3. At a certain instant, the particle of mass m has the position and velocity shown in the figure, and it is acted upon by the force F . Determine its angular momentum about point O and the time rate of change this angular momentum. Solution. The position and velocity vectors are r v = = ai + bj + ck vj 2 (13) (14) z F m v O a b c y x Figure 3: Problem 3 The angular momentum H O can be calculated as H = = r × mv (ai + bj + ck) × mvj = mv(−ci + ak) (15) When F = F k, the time rate of change of H O is Ḣ O = r×F = F (bi − aj) 3 (16)