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AP Physics 1 : Linear Motion and Momentum

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Example Questions

Example Question #31 : Impulse And Momentum

Conservation of momentum, $\Delta \vec{p} = 0$ , is true for which scenario?

Possible Answers:

Momentum is conserved if all forces, $\sum \vec{F}$ , are conservative.

Momentum is conserved only in one-dimensional collisions.

Momentum is conserved if no net external force, $\vec{F}_{ext}$ , is present.

Momentum is always conserved.

Correct answer:

Momentum is conserved if no net external force, $\vec{F}_{ext}$ , is present.

Explanation:

Momentum is conserved in isolated systems, (i.e. there are no net external forces.)

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Example Question #33 : Impulse And Momentum

Determine the momentum of a baseball of mass 145g moving at $44\frac{m}{s}$ .

Possible Answers:

140J

254J

None of these

361J

395J

Correct answer:

140J

Explanation:

Using the equation for kinetic energy

KE=.5mv^2

Converting to and plugging in values:

KE=.5*.145*44^2

KE=140J

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Example Question #34 : Impulse And Momentum

Two cars of mass 1450kg hit head on, each traveling at $30\frac{m}{s}$ . They then bounce back at $2.5\frac{m}{s}$ If the collision took .25 s , determine the magnitude of force experienced by each car.

Possible Answers:

95500N

Impossible to determine

188500N

188N

None of these

Correct answer:

188500N

Explanation:

Using momentum principle:

$P_i+F\Delta t=P_f$

Since the cars have identical masses and experience the same change in velocity, the forces on them will be identical.

$m_iv_i+F\Delta t=m_fv_f$

Plugging in values:

1450*30+F*.25=1450*-2.5

Solving for :

F=-188500N

|F|=188500N

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Example Question #37 : Impulse And Momentum

Two cars of mass 1450kg hit head on, each traveling at $30\frac{m}{s}$ . They then bounce back at $2.5\frac{m}{s}$ If the collision took .25 s , determine the force experienced by a driver of mass 95kg in one of the cars.

Possible Answers:

9.85*10^4N

2.30*10^5N

Impossible to determine

None of these

1.24*10^5N

Correct answer:

1.24*10^5N

Explanation:

Using momentum principle:

$P_i+F\Delta t=P_f$

Since the cars have identical masses and experience the same change in velocity, the forces on them will be identical.

$m_iv_i+F\Delta t=m_fv_f$

Plugging in values:

1450*30+F*.25=1450*-2.5

Solving for :

F=-188500N

|F|=188500N

Now, determine the acceleration of the car:

|F|=188500N=1450*a

$a=130\frac{m}{s^2}$

Now, determine the force experienced by the driver who was in the car, and thus accelerated with the car.

$F=m*a=95*130\frac{m}{s^2}$

F=1.24*10^5N

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Example Question #33 : Impulse And Momentum

When firefighters rescue a person from a burning building several stories high, they're prepared with a padded landing for when the person jumps down. As this person falls, they build up greater and greater momentum. How does the padded landing bring this falling person safely to rest?

Possible Answers:

It decreases the force of the collision by decreasing the time of the collision

It increases the force of the collision by decreasing the time of the collision

It increases the force of the collision by increasing the time of the collision

It decreases the force of the collision by increasing the time of the collision

Correct answer:

It decreases the force of the collision by increasing the time of the collision

Explanation:

For this question, we're asked how a padded landing helps a person fall safely from a large height. To do so, we need to consider how the landing interacts with the falling person in changing their momentum.

Let's first remember that a change in an object's momentum is called impulse, which can be written with the following expression.

$I=m\Delta v$

What's more, the units can be rearranged in such a way as to express impulse in terms of average force and time of the collision.

$I=m\Delta v=F_{avg}\Delta t$

Next, we can rearrange the above expression to make it easier to see how force and time of collision are related.

$F_{avg}=\frac{m \Delta v}{\Delta t}$

As we can see in the above expression, the average force of collision is inversely proportional to the time of the collision. In other words, the longer the collision lasts, the less the average force will be.

Relating this to the padded landing example, it's clear that the landing helps to reduce a person's fall by reducing the average force of the collision between the person and the landing. Moreover, the amount of time the collision takes to happen is increased.

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Example Question #41 : Impulse And Momentum

Deep in space Object has mass 350kg and is initially traveling with velocity $<30,50>\frac{m}{s}$ . At t=0 , it collides with Object , which has mass 500kg and is initially motionless. The two objects stick together.

Determine the initial momentum of the system.

Possible Answers:

$<17500,10500>\frac{kg*m}{s}$

$<17500,17500>\frac{kg*m}{s}$

Impossible to determine

$<10500,17500>\frac{kg*m}{s}$

None of these

Correct answer:

$<10500,17500>\frac{kg*m}{s}$

Explanation:

Using

$\overrightarrow{p}=m\overrightarrow{v}$

Plugging in values:

$\overrightarrow{p}=350kg*<30,50>\frac{m}{s}$

$\overrightarrow{p}=<10500,17500>\frac{kg*m}{s}$

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Example Question #41 : Impulse And Momentum

Determine the impulse experienced by Object

Possible Answers:

None of these

$<6200,10300>\frac{kg*m}{s}$

$<4449,39080>\frac{kg*m}{s}$

$<10300,6200>\frac{kg*m}{s}$

$<4340,7210>\frac{kg*m}{s}$

Correct answer:

$<6200,10300>\frac{kg*m}{s}$

Explanation:

Using

$\overrightarrow{p}=m\overrightarrow{v}$

Plugging in values:

$\overrightarrow{p}=350kg*<30,50>\frac{m}{s}$

$\overrightarrow{p}=<10500,17500>\frac{kg*m}{s}$

The momentum will be the same in the final state, so again using

$\overrightarrow{p}=m\overrightarrow{v}$

Solving for velocity:

$\frac{\overrightarrow{p}}{m}=\overrightarrow{v}$

Plugging in values (the total mass is equal to the combined masses:

$\frac{<10500,17500>}{850}=\overrightarrow{v}$

$\overrightarrow{v_f}=<12.4,20.6>\frac{m}{s}$

Definition of impulse:

$J=\Delta \overrightarrow{p}$

$J=m\Delta\overrightarrow{v}$

J=500(<12.4,20.6>-<0,0>)

$J=<6200,10300>\frac{kg*m}{s}$

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Example Question #43 : Impulse And Momentum

A ball of mass 30g is thrown at a target. The ball strikes with a velocity of $20\frac{m}{s}$ and bounces back with equal magnitude. Determine the magnitude of impulse experienced by the ball.

Possible Answers:

$1.2\frac{kg\cdot m}{s}$

$6.8\frac{kg\cdot m}{s}$

$3.6\frac{kg\cdot m}{s}$

$5.5\frac{kg\cdot m}{s}$

$7.7\frac{kg\cdot m}{s}$

Correct answer:

$1.2\frac{kg\cdot m}{s}$

Explanation:

Impulse is defined as change in momentum:

$\Delta P= P_f-P_i$

Using

P=mv

Combining equations:

$\Delta P= mv_f-mv_i$

Plugging in values:

$\Delta P= .030*-20-.030*20$

$|\Delta P| =1.2\frac{kg*m}{s}$

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Example Question #44 : Impulse And Momentum

Frictionless cart is traveling at $<10,15>\frac{m}{s}$ when it hits identical frictionless cart which was previously motionless. After the collision, cart is traveling at $<1,1.5>\frac{m}{s}$ . Determine the final velocity of cart .

Possible Answers:

$<18,27>\frac{m}{s}$

None of these

$<19,3.5>\frac{m}{s}$

$<9,13.5>\frac{m}{s}$

$<0,0>\frac{m}{s^2}$

Correct answer:

$<9,13.5>\frac{m}{s}$

Explanation:

Using conservation of momentum:

Pi=P_f

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Solving for $v_{2f}$

$\frac{m_1v_{1i}+m_2v_{2i}-m_1v_{1f}}{m_1}=m_2v_{2f}$

Using m_1=m_2 (since the carts are identical):

$v_{1i}+v_{2i}-v_{1f}=v_{2f}$

Plugging in values:

<10,15>+<0,0>-<1,1.5>=<9,13.5>

$v_{2f}=<9,13.5>\frac{m}{s}$

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Example Question #42 : Impulse And Momentum

During time period , a rocket ship deep in space of mass 2.55*10^5kg travels from <0,10>km to <5,5>km . During time period , the rocket fires. During time period , the rocket travels from <-15,0>km to <-30,-5>km .

Time periods , , and all took 10seconds.

Determine the impulse during time period .

Possible Answers:

None of these

$<5.1*10^8,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$\Delta\overrightarrow{P}=<0,5.1*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$\Delta\overrightarrow{P}=<0,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$\Delta\overrightarrow{P}=<5.1*10^8,5.1*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Correct answer:

$<5.1*10^8,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Explanation:

Finding initial momentum:

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<5000-0,5000-10000>}{10}$

$\overrightarrow{P_i}=<1.275*10^8,-1.275*10^8>\frac{kg*m}{s}$

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Converting to and plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<-30000+15000,-5000-0>}{10}$

$\overrightarrow{P}_f=<-3.825*10^8,-1.275*10^8>\frac{kg*m}{s}$

Using

$Impulse=\Delta\overrightarrow{P}=\overrightarrow{P_f}-\overrightarrow{P_i}$ Plugging in values:

$\overrightarrow{\Delta P} =<-3.825*10^8,-1.275*10^8>-<1.275*10^8,-1.275*10^8>$

$\Delta\overrightarrow{P}=<5.1*10^8,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

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AP Physics 1 : Linear Motion and Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #31 : Impulse And Momentum

Example Question #33 : Impulse And Momentum

Example Question #34 : Impulse And Momentum

Example Question #37 : Impulse And Momentum

Example Question #33 : Impulse And Momentum

Example Question #41 : Impulse And Momentum

Example Question #41 : Impulse And Momentum

Example Question #43 : Impulse And Momentum

Example Question #44 : Impulse And Momentum

Example Question #42 : Impulse And Momentum

All AP Physics 1 Resources

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