Practice 6C: | 1 | 2 | 3 | Go up
Momentum and Impulse - by Stephanie Culnane, 2001

1. How long would it take the car in Sample Problem 6C to come to a stop from 20.0 m/s to the west?  How far would the car move before stopping?  Assume a constant acceleration. 

Here's what you know, m = 2250 kg, u = 20 m/s, v = 0 m/s, and F = 8450 N.  Use the formula Dt = Dp/F = (mv_f - mv_i)/F to find the time.  Plug in Dt =  ( (2250 * 20)(2250 * 0))/8450 N = 5.33 s. To find the distance use the formula Dx = ¹/₂(v_i + v_f)Dt.  Plug in Dx = 1/2(20 + 0)(5.33).  Dx = 53.3 m to the west.
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2. A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20.0 m/s by a 6250 N braking force acting opposite the car's motion.  Use the impulse-momentum theorem  to answer the following questions:

            a.  What is the car's velocity after 2.50 s?

            b.  How far does the car move during 2.50 s?

            c.  How long does it take the car to come to a complete stop?

Here's what you know, m = 2500 kg, u = 20.0 m/s, F = -6250 N, and t = 2.50 s.  

a.  Use the formula FDt = mv_f - mv_i.  Plug in (-6250 N)(2.50s) = (2500 kg)(v) - (2500kg)(20 m/s).  v = 13.75 m/s

b.   Use the formula Dx = ¹/₂(v_i + v_f)Dt.  Plug in Dx = 1/2(20 m/s + 13.75 m/s)(2.50 s).  Dx = 42.19 m.
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3.  Assume that the car in Sample Problem 6C has a mass of 3250 kg.

a.  How much force would be required to cause the same acceleration as in item 1?  Use the impulse-momentum theorem.

b.  How far would the car move before stopping?

Here's what you know, m = 3250 kg, t = 5.33 s, and vf = 20 m/s 

a. Use the formula Dt = (mv_f - mv_i)/F. Plug in 5.33 s = (3250 kg)(20 m/s) - 0/F. F = 12,195 N = 1.22 * 10^4 N. 

b. Use the formula Dx = 1/2(v_f - v_i)Dt. Plug in Dx = 1/2(20 m/s + 0)(5.33s). Dx = 53.3 m to the west.
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