Question

Sometimes, in an intense battle, gunfire is so intense that bullets from opposite sides collide in midair. Suppose that one (with mass M = 5.12 g moving to the right at a speed V = 227 m/s directed 21.3◦ above the horizontal) collides and fuses with another with mass m = 3.05 g moving to the left at a speed v = 282 m/s directed 15.4◦ above the horizontal.

(a) What is the magnitude (m/s) of their common velocity immediately after the collision?

(b) What is the direction (degrees) of their common velocity immediately after the collision? (Measure this angle from the horizontal.)

(c) What fraction of the original kinetic energy was lost in the collision?

Answer 1

Ve 1213 15.4 Mir 3.05 g ma M, = 5.129 =227"/s . Momentum a Conservation of (M,4 m2) ✓ m. Vilces 21.3 € + Sin 21-39 ) + , v, -- -66154î + sin15:49 5.124 2271 0021.3 î + Sin 21-39) (0) 15.4 Î + Sin 15.49) (5-12 + Des î x 79.63 ms. of 3:05x282 j 31 > mis 312+790632 a > lvl= 85.45 85.45 mls 79.63 b) tano a horizontal 68.73 from os

Initial KG I m, vt + 2 mu? 5-12x173(227) + 3-05410"}(282)) = KES 1 2 253.2 J KE 17 1 (mens) v 올 ( 5-12 43-05) 210 *(8sous) s 99.827 I kily 223.45 253.2 - 29.827 lost Energy lost fraction = original Ke 223.4 253.2 0.882