D. Momentum Homework Part I
6.1 – 6.2
6.1 Linear Momentum:
1, 3, 4, 7, 10, 13, 15, 16, 17, 18, 21
10: Remember vo is negative and v is positive because the ball is changing directions. It could also be the other way around. It doesn’t matter.
13: a. v = vo + gt where vo = 0m/s and t = 0.75s. b. v2 = vo2 + 2gy where vo = 0 m/s,
y = 10 m and you solve for v.
15: The two runners could be running in the same direction OR running AT each other. The net momentum is an ABSOLUTE value sum.
16: Easy, just do the resulting vector. DON’T resolve the VECTOR!
17: Now you need to resolve the vectors since the rebound angle doesn’t equal the impact angle. The incoming ball is 60o west of south and the rebound ball is 50o east of south. Look at figure 6.26 to visualize it.
18: Can it get any easier than this?
21: 1 Kg = 2.2 lbs – change into lbs first! vo = v but opposite in direction. Therefore,
DP = m(2vo). B. v will NOT be 4.5 m/s. To determine v, use v2 = vo2 – 2gy where vo = +4.50 m/s and y = -0.25 m. You’ll find v to be 5.01 m/s.
Then, DP = mDv = 54.5 kg * (-5.01m/s – 4.50m/s)
22, 23, 25, 26, 27, 29, 34, 35
23: Shorter contact time will result in larger force if all other factors (m, vo and v) remain the same.
25: Increase contact time decreases force.
26: I = FDt = m*a*t = m*v (because a * t = v)
34: a. Impulse will equal the area under the curve. The max force on that curve is 900N.
b. Average F = impulse/Dt. c. FDt = DP = mv – mvo Where vo = -6.0 m/s.
35: find impact velocity with v = square root of 2gh.
Then, Average FDt = DP = mv – mvo where vo = 0m/s
4: a. 1.5 x 103 kg m/s b. zero
10: 5.88 kg m/s in the direction opposite to vo
16: -3 N s
18: 16 N
26: 15 m/s
34: a. 77 N s b. 5.5 x 102N c. 20 m/s