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S E C T I O N 9 . 4 • Two-Dimensional Collisions 269 Because the total momentum in the x direction is con- Similarly, the total initial momentum of the system in the y direction is that of the van, and the magnitude of this momen- 4 kg ) m/s. Applying conservation of momentum to the y direction, we have If we divide Equation (2) by Equation (1), we obtain (2) 5.00 * 10 4 kg)m/s ! (4 000 kg)v f
sin & " p yi ! " p yf (1) 3.75 * 10 4 kg)m/s ! (4 000 kg)v f
cos & " p xf ! (4 000 kg)v f cos & Example 9.11 Proton–Proton Collision A proton collides elastically with another proton that is ini- 5 m/s and makes a glancing collision with the sec- ond proton, as in Figure 9.13. (At close separations, the Solution The pair of protons is an isolated system. Both 1 ! m 2 , & ! 37.0°, and we are given that v 1i ! 3.50 * 10 5 m/s, Equations 9.24, 9.25, and 9.26 become (1) (3) We rewrite Equations (1) and (2) as follows: Now we square these two equations and add them: v 2f 2 ! 1.23 * 10 11 " (5.59 * 10 5 )v 1f # v 1f 2
# v 1f 2 cos 2 37.0, # v 1f 2 sin 2 37.0, ! 1.23 * 10 11 m 2 /s 2 " (7.00 * 10 5 m/s)v 1f cos 37.0, v 2f 2 cos 2 + # v 2f 2 sin 2 + v 2f sin + ! v 1f sin 37.0, v 2f cos + ! 3.50 * 10 5 m/s "
v 1f cos 37.0, ! 1.23 * 10 11 m 2 /s 2 v 1f 2 # v 2f 2 ! (3.50 * 10 5 m/s) 2 (2) v 1f sin 37.0, " v 2f sin + ! 0 v 1f cos 37, # v 2f cos + ! 3.50 * 10 5 m/s Substituting into Equation (3) gives One possibility for the solution of this equation is v 1f ! 0, which corresponds to a head-on collision—the first proton From Equation (3), and from Equation (2), It is interesting to note that & # + ! 90°. This result is not 53.0, ! + ! sin " 1 # v 1f sin 37.0, v 2f $ ! sin " 1 # (2.80 * 10 5 ) sin 37.0, 2.12 * 10 5 $ ! 2.12 * 10 5 m/s v 2f ! √ 1.23 * 10 11 " v 1f 2 ! √ 1.23 * 10 11 " (2.80 * 10 5 ) 2 2.80 * 10 5 m/s 2v 1f " 5.59 * 10 5 ! 0 9: v 1f ! 2v 1f 2 " (5.59 * 10 5 )v 1f ! (2v 1f " 5.59 * 10 5 )v 1f ! 0 ! 1.23 * 10 11 v 1f 2 # '1.23 * 10 11 " (5.59 * 10 5 )v 1f # v 1f 2
( & ! When this angle is substituted into Equation (2), the value f is It might be instructive for you to draw the momentum vec- 15.6 m/s v f ! 5.00 * 10 4 kg)m/s (4 000 kg)sin 53.1, ! 53.1,
sin & cos & ! tan & ! 5.00 * 10 4 3.75 * 10 4 ! 1.33 Example 9.12 Billiard Ball Collision In a game of billiards, a player wishes to sink a target ball in Solution Let ball 1 be the cue ball and ball 2 be the target But m 1 ! m 2 ! m, so that 1 2 m 1 v 1i 2 ! 1 2 m 1 v 1f 2 # 1 2 m 2 v 2f 2 |