Q: A 2000 kg pickup truck is moving at $$10 \, \mathrm{m/s}$$ east. What is its momentum?

$$p = 2000 \, \mathrm{kg} \times 10 \, \mathrm{m/s} = 20{,}000 \, \mathrm{kg} \cdot \mathrm{m/s}$$ \text{ east}

Q: A 1200 kg car traveling west at $$15 \, \mathrm{m/s}$$ crashes into a utility pole and comes to a stop in $$0.30 \, \mathrm{s}. $$What is the average force exerted on the car during the collision?

$$F = \frac{\Delta p}{\Delta t} = \frac{1200 \times (0 - 15)}{0.30} = -60{,}000 \, \mathrm{N}$$ \text{ (east)}

Q: A 1500 kg car traveling west slows from $$20.0 \, \mathrm{m/s} to$$ $$5.00 \, \mathrm{m/s}$$ under a force of $$8450 \, \mathrm{N}$$ east. How long does it take the car to decelerate?

$$t = \frac{m (v_f - v_i)}{F} = \frac{1500 \times (5.00 - 20.0)}{8450} = -2.66 \, \mathrm{s}$$

Q: A soccer ball of mass $$0.45 \, \mathrm{kg} is $$kicked and accelerates from rest to $$25 \, \mathrm{m/s} in$$ $$0.05 \, \mathrm{s}. $$What is the average force exerted on the ball?

$$F = \frac{\Delta p}{\Delta t} = \frac{0.45 \times 25}{0.05} = 225 \, \mathrm{N}$$

Q: A baseball of mass $$0.15 \, \mathrm{kg}$$ moving at $$40 \, \mathrm{m/s} is $$caught and brought to rest in $$0.02 \, \mathrm{s}. $$What is the magnitude of the impulse delivered to the ball?

$$J = \Delta p = m (v_f - v_i) = 0.15 \times (0 - 40) = -6.0 \, \mathrm{kg} \cdot \mathrm{m/s}$$

Question 1

Which of the following best defines linear momentum for a moving object?

Accepted Answer

The product of an object's mass and its velocity: $$p = m v$$

Question 2

A 2000 kg pickup truck is moving at $$10 \, \mathrm{m/s}$$ east. What is its momentum?

Accepted Answer

$$p = 2000 \, \mathrm{kg} 	imes 10 \, \mathrm{m/s} = 20{,}000 \, \mathrm{kg} \cdot \mathrm{m/s}$$ 	ext{ east}

Question 3

Which equation correctly relates impulse to the change in momentum?

Accepted Answer

$$J = \Delta p = F \Delta t$$

Question 4

A 1200 kg car traveling west at $$15 \, \mathrm{m/s}$$ crashes into a utility pole and comes to a stop in $$0.30 \, \mathrm{s}. $$What is the average force exerted on the car during the collision?

Accepted Answer

$$F = \frac{\Delta p}{\Delta t} = \frac{1200 	imes (0 - 15)}{0.30} = -60{,}000 \, \mathrm{N}$$ 	ext{ (east)}

Question 5

Why do athletes 'follow through' in sports such as baseball or golf?

Accepted Answer

To increase the time over which force is applied, maximizing the impulse and thus the change in momentum.

Question 6

A 1500 kg car traveling west slows from $$20.0 \, \mathrm{m/s} to$$ $$5.00 \, \mathrm{m/s}$$ under a force of $$8450 \, \mathrm{N}$$ east. How long does it take the car to decelerate?

Accepted Answer

$$t = \frac{m (v_f - v_i)}{F} = \frac{1500 	imes (5.00 - 20.0)}{8450} = -2.66 \, \mathrm{s}$$

Question 7

During a car crash, why is it safer to have airbags that increase the time over which the force acts?

Accepted Answer

Increasing the time reduces the average force experienced by the passenger, according to $$F = \frac{\Delta p}{\Delta t}.$$

Question 8

Which of the following statements about impulse is correct?

Accepted Answer

Impulse is a vector quantity and has the same direction as the applied force.

Question 9

A soccer ball of mass $$0.45 \, \mathrm{kg} is $$kicked and accelerates from rest to $$25 \, \mathrm{m/s} in$$ $$0.05 \, \mathrm{s}. $$What is the average force exerted on the ball?

Accepted Answer

$$F = \frac{\Delta p}{\Delta t} = \frac{0.45 	imes 25}{0.05} = 225 \, \mathrm{N}$$

Question 10

Which of the following is a real-world application of the impulse-momentum theorem?

Accepted Answer

Engineers use it to design safe stopping distances for vehicles.

Question 11

A baseball of mass $$0.15 \, \mathrm{kg}$$ moving at $$40 \, \mathrm{m/s} is $$caught and brought to rest in $$0.02 \, \mathrm{s}. $$What is the magnitude of the impulse delivered to the ball?

Accepted Answer

$$J = \Delta p = m (v_f - v_i) = 0.15 	imes (0 - 40) = -6.0 \, \mathrm{kg} \cdot \mathrm{m/s}$$

Question 12

A 1000 kg car traveling at $$18 \, \mathrm{m/s}$$ comes to a stop in $$4.0 \, \mathrm{s}. $$What is the average force required to stop the car?

Accepted Answer

$$F = \frac{\Delta p}{\Delta t} = \frac{1000 	imes (0 - 18)}{4.0} = -4{,}500 \, \mathrm{N}$$

Question 13

If a force of $$500 \, \mathrm{N}$$ acts on a $$2.0 \, \mathrm{kg}$$ object for $$0.10 \, \mathrm{s}$$, what is the change in velocity of the object?

Accepted Answer

$$\Delta v = \frac{F \Delta t}{m} = \frac{500 	imes 0.10}{2.0} = 25 \, \mathrm{m/s}$$

Question 14

A 0.2 kg tennis ball moving at $$30 \, \mathrm{m/s} is $$hit back in the opposite direction at $$40 \, \mathrm{m/s} in$$ $$0.05 \, \mathrm{s}. $$What is the average force exerted by the racket?

Accepted Answer

$$F = \frac{m (v_f - v_i)}{\Delta t} = \frac{0.2 	imes (-40 - 30)}{0.05} = -280 \, \mathrm{N}$$

Question 15

Which of the following best describes the impulse-momentum theorem?

Accepted Answer

The change in momentum of an object is equal to the impulse applied to it: $$\Delta p = F \Delta t$$

Question 16

A 900 kg car traveling at $$25 \, \mathrm{m/s}$$ east comes to a stop in $$2.0 \, \mathrm{s}. $$What is the magnitude and direction of the average force exerted on the car?

Accepted Answer

$$F = \frac{900 	imes (0 - 25)}{2.0} = -11{,}250 \, \mathrm{N}$$ 	ext{ (east)}

Question 17

A 2.0 kg object initially at rest is acted upon by a constant force of $$10 \, \mathrm{N}$$ for $$0.5 \, \mathrm{s}. $$What is its final velocity?

Accepted Answer

$$v_f = \frac{F \Delta t}{m} = \frac{10 	imes 0.5}{2.0} = 2.5 \, \mathrm{m/s}$$

Question 18

A 1000 kg car traveling at $$20 \, \mathrm{m/s}$$ slows to $$5 \, \mathrm{m/s}$$ over a distance of $$50 \, \mathrm{m}. $$What is the average net force acting on the car?

Accepted Answer

$$F = m a$$, 	ext{ where } a = \frac{v_$$f^2 - v$$_$$i^2$$}{2d} = \frac{$$5^2$$ - $$20^2$$}{2 	imes 50} = -3.75 \, \mathrm{$$m/s^2$$}$$, 	ext{ so } F = 1000 	imes (-3.75) = -3{,}750 \, \mathrm{N}$$

Momentum and Impulse: Concepts, Formulas, and Applications

Study Guide - Practice Questions

Study Guide - Flashcards