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AP Physics 1 : AP Physics 1

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Example Questions

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.

$I=m\Delta v$ $\displaystyle I=m\Delta v$

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.

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

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

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

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

Determine the initial momentum of the system.

Possible Answers:

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

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

Impossible to determine

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

None of these

Correct answer:

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

Explanation:

Using

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

Plugging in values:

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

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

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Example Question #41 : Impulse And Momentum

Determine the impulse experienced by Object $\displaystyle B$

Possible Answers:

None of these

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

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

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

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

Correct answer:

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

Explanation:

Using

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

Plugging in values:

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

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

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

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

Solving for velocity:

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

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

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

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

Definition of impulse:

$J=\Delta \overrightarrow{p}$ $\displaystyle J=\Delta \overrightarrow{p}$

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

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

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

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Example Question #43 : Impulse And Momentum

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Impulse is defined as change in momentum:

$\Delta P= P_f-P_i$ $\displaystyle \Delta P= P_f-P_i$

Using

P=mv $\displaystyle P=mv$

Combining equations:

$\Delta P= mv_f-mv_i$ $\displaystyle \Delta P= mv_f-mv_i$

Plugging in values:

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

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

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Example Question #221 : Linear Motion And Momentum

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

Possible Answers:

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

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

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

None of these

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

Correct answer:

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

Explanation:

Using conservation of momentum:

Pi=P_f $\displaystyle Pi=P_f$

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

Solving for $v_{2f}$ $\displaystyle v_{2f}$

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

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

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

Plugging in values:

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

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

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Example Question #42 : Impulse And Momentum

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

Time periods $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ all took $\displaystyle 10seconds.$

Determine the impulse during time period $\displaystyle B$.

Possible Answers:

None of these

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

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

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

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

Correct answer:

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

Explanation:

Finding initial momentum:

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

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

Combining equations:

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

Plugging in values:

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

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

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

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

Combining equations:

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

Converting $\displaystyle km$ to $\displaystyle m$ and plugging in values:

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

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

Using

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

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

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

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Example Question #231 : Linear Motion And Momentum

Ty throws a 0.10kg $\displaystyle 0.10kg$ snowball directly at his brother who also throws a snowball directly at Ty with a mass of 0.20kg $\displaystyle 0.20kg$, because he dipped it in a bucket of water. Ty throws his snowball at $20.0\frac{m}{s}$ $\displaystyle 20.0\frac{m}{s}$ and his brother throws his at $-15\frac{m}{s}$ $\displaystyle -15\frac{m}{s}$ and the two snowballs make perfect contact in mid flight, causing a perfectly inelastic collision. Neglecting air resistance, what is the combined final speed of the snowballs and their direction following the collision?

Possible Answers:

$6.6\frac{m}{s}$ $\displaystyle 6.6\frac{m}{s}$; positive x-direction

$-3.3\frac{m}{s}$ $\displaystyle -3.3\frac{m}{s}$; negative x-direction

$-6.6\frac{m}{s}$ $\displaystyle -6.6\frac{m}{s}$; negative x-direction

$3.3\frac{m}{s}$ $\displaystyle 3.3\frac{m}{s}$; positive x-direction

Correct answer:

$-3.3\frac{m}{s}$ $\displaystyle -3.3\frac{m}{s}$; negative x-direction

Explanation:

Use the conservation of momentum for inelastic collisions (objects that stick together).

$m_{Ty}(v_{Ty})+m_{brother}(v_{brother})=(m_{Ty}+m_{brother})v_{final}$ $\displaystyle m_{Ty}(v_{Ty})+m_{brother}(v_{brother})=(m_{Ty}+m_{brother})v_{final}$

Plug in and solve for the final velocity.

$0.1Kg(20\frac{m}{s})+0.2Kg(-15\frac{m}{s})=(0.1Kg+0.2Kg)v_{final}$ $\displaystyle 0.1Kg(20\frac{m}{s})+0.2Kg(-15\frac{m}{s})=(0.1Kg+0.2Kg)v_{final}$

$2j+(-3j)=(0.3Kg)v_{final}$ $\displaystyle 2j+(-3j)=(0.3Kg)v_{final}$

$v_{final}=-3.33\frac{m}{s}$ $\displaystyle v_{final}=-3.33\frac{m}{s}$

The final velocity of the combined snowballs is $-3.3\frac{m}{s}$ $\displaystyle -3.3\frac{m}{s}$ in the negative x-direction.

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Example Question #47 : Impulse And Momentum

On a toy car set, two children apply a piece of double sided tape to the front of two toy cars so that when they collide they will stick together causing a perfectly inelastic collision. The two identical cars have a combined mass of 0.075kg $\displaystyle 0.075kg$ and their final combined speed was $2.60\frac{m}{s}$ $\displaystyle 2.60\frac{m}{s}$ in the rightward direction. If car "A" was shot to the right with a speed measured at $17.0\frac{m}{s}$ $\displaystyle 17.0\frac{m}{s}$ right before the collision . At what speed was the other car traveling right before the collision?

Possible Answers:

$-6.1\frac{m}{s}$ $\displaystyle -6.1\frac{m}{s}$

$-11.8\frac{m}{s}$ $\displaystyle -11.8\frac{m}{s}$

$-18.4\frac{m}{s}$ $\displaystyle -18.4\frac{m}{s}$

$-24.5\frac{m}{s}$ $\displaystyle -24.5\frac{m}{s}$

Correct answer:

$-11.8\frac{m}{s}$ $\displaystyle -11.8\frac{m}{s}$

Explanation:

Use the conservation of momentum, perfectly inelastic collision.

$m_{A}(v_{A})+m_{B}(v_{B})=(m_{A}+m_{B})v_{final}$ $\displaystyle m_{A}(v_{A})+m_{B}(v_{B})=(m_{A}+m_{B})v_{final}$

Plug in and solve for the speed of car B.

$0.0375Kg(17\frac{m}{s})+0.0375Kg(v_{B})=(0.075Kg)2.6\frac{m}{s}$ $\displaystyle 0.0375Kg(17\frac{m}{s})+0.0375Kg(v_{B})=(0.075Kg)2.6\frac{m}{s}$

$0.6375j+0.0375Kg(v_{B})=0.19j$ $\displaystyle 0.6375j+0.0375Kg(v_{B})=0.19j$

$v_{B}=-11.8\frac{m}{s}$ $\displaystyle v_{B}=-11.8\frac{m}{s}$

The velocity of car B right before the collision was $-11.8\frac{m}{s}$ $\displaystyle -11.8\frac{m}{s}$.

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Example Question #48 : Impulse And Momentum

A spaceship is stationary deep in space with it’s rocket broken. The crew decides to propel it to earth by throwing tennis balls out the back window. If they throw one tennis ball every second at $30 \frac{m}{s}$ $\displaystyle 30 \frac{m}{s}$, how many balls will it take the ship to reach $1.1*10^4 \frac{m}{s}$ $\displaystyle 1.1*10^4 \frac{m}{s}$? Each tennis ball has a mass of 50 g $\displaystyle 50 g$ and the ship has a mass of $\displaystyle 7.0*10^6kg$.

Possible Answers:

None of these

$5.13*10^{10}balls$ $\displaystyle 5.13*10^{10}balls$

$6.26*10^{14}balls$ $\displaystyle 6.26*10^{14}balls$

$8.44*10^{7}balls$ $\displaystyle 8.44*10^{7}balls$

$3.95*10^{18}balls$ $\displaystyle 3.95*10^{18}balls$

Correct answer:

$5.13*10^{10}balls$ $\displaystyle 5.13*10^{10}balls$

Explanation:

The net momentum of the system needs to stay constant, that is, any momentum given to the tennis balls, the space ship must gain in the opposite direction.

$\displaystyle m_sv_s+n_bm_bv_b=0$

Where

m_s $\displaystyle m_s$ is the mass of the ship.

v_s $\displaystyle v_s$ is the velocity of the ship

n_b $\displaystyle n_b$ is the number of balls thrown

m_b $\displaystyle m_b$ is the mass of a ball

v_b $\displaystyle v_b$ is the velocity of a thrown ball

Solving for $\displaystyle n_b$

$n_b=\frac{m_sv_s}{m_b(-v_b)}$ $\displaystyle n_b=\frac{m_sv_s}{m_b(-v_b)}$

Since the velocities are given as magnitudes, the negative sign can be ignored

$n_b=\frac{m_sv_s}{m_b(v_b)}$ $\displaystyle n_b=\frac{m_sv_s}{m_b(v_b)}$

Plugging in values

$n_b=\frac{7*10^6*1.1*10^4}{.050(30)}$ $\displaystyle n_b=\frac{7*10^6*1.1*10^4}{.050(30)}$

$n_b=5.13*10^{10}$ $\displaystyle n_b=5.13*10^{10}$

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Example Question #41 : Impulse And Momentum

A bullet with mass $\displaystyle 5g$ is shot with an initial kinetic energy of 2.5kJ $\displaystyle 2.5kJ$. What was the impulse on the bullet as it was fired?

Possible Answers:

$20N\cdot s$ $\displaystyle 20N\cdot s$

$10N\cdot s$ $\displaystyle 10N\cdot s$

$25N\cdot s$ $\displaystyle 25N\cdot s$

$5N\cdot s$ $\displaystyle 5N\cdot s$

$15N\cdot s$ $\displaystyle 15N\cdot s$

Correct answer:

$5N\cdot s$ $\displaystyle 5N\cdot s$

Explanation:

The expression for impulse is the following:

$I = F\Delta t=m\Delta v$ $\displaystyle I = F\Delta t=m\Delta v$

We are given mass and initial kinetic energy, so we can also calculate the velocity of the bullet. Therefore, we will go with the second for of the expression:

$I = m\Delta v$ $\displaystyle I = m\Delta v$

Finding an expression for velocity, we get:

$K= \frac{1}{2}mv^2$ $\displaystyle K= \frac{1}{2}mv^2$

Rearranging for velocity, we get:

$v = \sqrt{\frac{2K}{m}}$ $\displaystyle v = \sqrt{\frac{2K}{m}}$

Plugging this in, we get:
$I = m\sqrt{\frac{2K}{m}}=\sqrt{2Km}$ $\displaystyle I = m\sqrt{\frac{2K}{m}}=\sqrt{2Km}$

Plugging in our values, we get:

$I = 5N\cdot s$ $\displaystyle I = 5N\cdot s$

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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

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 #221 : Linear Motion And Momentum

Example Question #42 : Impulse And Momentum

Example Question #231 : Linear Motion And Momentum

Example Question #47 : Impulse And Momentum

Example Question #48 : Impulse And Momentum

Example Question #41 : Impulse And Momentum

All AP Physics 1 Resources

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