BC Science Physics 12: Chapter 3

Page 20

Sample Problem 3.3.2 — Collision Between Two Objects: Vector Components (Continued) What to Think About 2. In Figure 3.3.2, the resultant momentum   ( pR = pio ) was drawn first. Then the directions   of pif and psf were constructed. As you can see, it is then a simple task to complete the momentum vector triangle.

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  3. After the collision, pR = pio = 0.10 kg⋅m/s in the direction shown. Use trigonometry to   solve for the magnitudes of pif and psf .

Figure 3.3.2   pif = pR cos 60 o  pif = ( 0.10 kg • m s)( 0.500 )  pif = 0.050 kg • m s  pif 0.050 kg • m s  = v if = = 1.0 m s 0.050 kg mi

Challenge: Can you see a shorter way to solve this particular problem?

86 Chapter 3 Momentum and Energy

Similarly,   psf = pR sin 60 o  psf = ( 0.10 kg • m s)( 0.866 )  psf = 0.087 kg • m s  psf 0.087 kg • m s  = v sf = = 1.7 m s 0.050 kg ms

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