Practice 6B: | 1 | 2
| 3 | 4 | Go up

**Momentum and Impulse - by Stephanie **C**ulnane,
2001**

**1.** ** A 0.50 kg football is thrown with a velocity of 15 m/s to the right. A stationary reciever catches the ball and brings it to rest in 0.020 s. What is
the force exerted on the reciever?**

Here's what you know, m = 0.50 kg, v = 15 m/s, and Dt = .020 s. Use the formula
FDt = p and plug in mv for p so that FDt = mv. Plug in F(.020 s)

= (0.50 kg)(15 m/s). F = 375 N = 3.8 * 10^2 N to the right.

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**2. An 82 kg man drops from rest on a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What
force does the water exert on him?
**

Here's what you know, m = 82 kg, s = 3.0 m, t = 0.55s. Use the formula v

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Here's what you know, m = 0.40 kg, u = 18 m/s, and v = -22 m/s. Use the formula FDt = mv

m/s) - (0.40kg)(-22 m/s). p = 16.0 kgm/s south.

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a. What is the velocity of the object at the end of this interval?

b. At the end of this interval, a constant force of 4.00 N to the left is applied for 3.00 s. What is the velocity at the end

of the 3.00 s?

Here's what you know, m = 0.50 kg, F = 3.00 N, and t = 1.50 s.

a. Use the formula FDt = p, plug in mv for p. Plug in (3.00N)(1.50 s) = (0.50 kg)(v). v =
+9 m/s.

Here's what you know, m = 0.50 kg, F = 4.00 N, and t = 3.00 s.

b. Use the formula FDt = p, plug in mv for p. Plug in
(Force is negative as it is to the left) (-4.00 N)(3.00s) = (0.50 kg)(v). v = -24 m/s.
- but this is the change in velocity - final is +9.00 m/s - 24 m/s = -15 m/s

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