Question

10.0 points 003 A 7.65 g bullet is stopped in a 6.4 kg block of wood. The speed of the bullet-plus-wood combination immediately after the collision is 0.851 m/s. What was the original speed of the bullet? Answer in units of m/s. 10.0 points 004 Two cars, one of mass 1100 kg, and the second of mass 2100 kg, are moving at right angles to each other when they collide and stick to- gether. The initial velocity of the first car is 13 m/s in the positive x direction and that of the second car is 17 m/s in the positive y direction What is the magnitude of the velocity of the wreckage of the two cars immediately after the collision? Answer in units of m/s. 10.0 points 005 Two ice skaters approach each other at right angles. and travels in the +x direction at 0.912 m/s. Skater B has a mass of 40.5 kg and is moving in the +y direction at 0.61 m/s. They collide and cling together. Skater A has a mass of 34.6 kg

Answer 1

(0.03) From the momentum conservation m4, + m, 4, = (m. + m2) v The initial velocity of the bullet is, (m + m2 )v-m, u = (7.65x10° +6.4)(0.851)-(6.4x100) 7.65x10 = 712.7987 m/s =713 m/s (0.04) The momentum before collision Pa = 1100x13a, ve=134 Po = 2100x17a, Vo = 17a, The total momentum before collision is, Pi = 14300a, +35700a, ---(1) The stick together, then total mass of wreckage is, P, =1100+ 2100 = 3200 kg The final momentum is, P, = 3200v; ---(2) Using conservation of momentum P: = p; 14300q, +35700a, = 32001 v= 14300q, +35700 a,) = 4.46875ą. +11.156a, Fl= (4.46875)* +(11.156) = 12.0 m/s