Practice 6C: | 1 | 2
| 3 | Go up

**Momentum and Impulse - by Stephanie **C**ulnane,
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/sb. 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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