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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #30 : Impulse And Momentum

A locomotive of mass 167*10^3kg traveling at $4\frac{km}{hr}$ couples to a motionless coach without applying the brakes. This took .5s . Determine the average force experienced by the coach.

Possible Answers:

105kN

207kN

302kN

382kN

507kN

Correct answer:

105kN

Explanation:

Use conservation of momentum

P_i=P_f

$m_{1i}v_{1i}+m_{2i}v_{2i}=m_{1,2f}v_{1,2f}$

Plug in values:

$(167*10^3)*4+(67*10^3)*0=(234*10^3)v_{1,2f}$

Solve for $v_{1,2f}$ :

$v_{1,2f}=\frac{167*10^3*4}{234*10^3}$

$v_{1,2f}=2.85\frac{km}{hr}$

Definition of impulse:

$J=\Delta P=m(v_f-v_i)$

Convert $\frac{km}{hr}$ to $\frac{m}{s}$

$2.85\frac{km}{hr}*\frac{1hr}{3600s}*\frac{1000m}{1km}=.79\frac{m}{s}$

Plug in values:

$J=\Delta P=67*10^3(.79-0)$

$J=52.3*10^3\frac{kg*m}{s}$

$J=F*\Delta t$

Plug in values

$52.3*10^3\frac{kg*m}{s}=F*.5seconds$

Solve for :

F=105kN

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Example Question #491 : Ap Physics 1

A locomotive of mass 167*10^3kg traveling at $4\frac{km}{hr}$ couples to a motionless passenger car without applying the brakes. Assuming no momentum losses, calculate the final velocity.

Possible Answers:

$3.33\frac{km}{hr}$

None of these

$1.05\frac{km}{hr}$

$2.85\frac{km}{hr}$

$5.95\frac{km}{hr}$

Correct answer:

$2.85\frac{km}{hr}$

Explanation:

Use conservation of momentum

P_i=P_f

$m_{1i}v_{1i}+m_{2i}v_{2i}=m_{1,2f}v_{1,2f}$

Plug in values:

$(167*10^3)*4+(67*10^3)*0=(234*10^3)v_{1,2f}$

Solve for $v_{1,2f}$ :

$v_{1,2f}=\frac{167*10^3*4}{234*10^3}$

$v_{1,2f}=2.85\frac{km}{hr}$

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Example Question #31 : Impulse And Momentum

A tennis player hits an incoming tennis ball of mass 50g and speed $40\frac{m}{s}$ right back at his opponent at a speed of $35\frac{m}{s}$ in the exact opposite direction. Suppose the collision between the racket and the tennis ball lasted for 5ms .

What is the magnitude of the average force exerted by the racket on the tennis ball?

Possible Answers:

750000N

750N

7500N

75N

0.75N

Correct answer:

750N

Explanation:

The average force exerted by the racket is related to the impulse by

$F\Delta t=I$

The impulse is related to the change in momentum by

$I=\Delta p=m\Delta v$

Equating the two expressions for impulse gives us

$F\Delta t=m\Delta v$

Simplify to get

$F=\frac{m\Delta v}{\Delta t}$

The change in velocity $\Delta v$ is given as

$\Delta v=v_f-v_i=-35\frac{m}{s}-40\frac{m}{s}=-75\frac{m}{s}$

Plugging in this, along with the given change in time and mass of the tennis ball gives us:

$F=\frac{(-75\frac{m}{s})(0.05kg)}{0.005s}=-750N$

Since the question asks for the magnitude of the average force, we get rid of the negative above to obtain F=750N

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Example Question #32 : Impulse And Momentum

A car takes to accelerate to $100\frac{km}{hr}$ from rest. Determine the average force experience by a driver of mass 100kg .

Possible Answers:

695N

1080N

395N

844N

788N

Correct answer:

695N

Explanation:

Use the equation for momentum:

$F_{net}*\Delta t=P_f-P_i$

$F_{net}*\Delta t=mv_f-mv_i$

Initial velocity was

$F_{net}*\Delta t=mv_f$

$\frac{100km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600s}=\frac{27.8m}{s}$

Plug in values:

$F_{net}*4=100*27.8$

$F_{net}=695N$

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Example Question #31 : Impulse And Momentum

A popular topic in early space exploration was how to safely return modules back to the surface of the earth. Early designs contained materials that could only withstand impulses of up to $4500 \frac{kg\cdot m}{s}$ before parts of the module became compromised. One such model of mass 500kg is approaching ocean waters and deploys its shoot, reducing its speed to $8 \frac{m}{s}$ . If the module decelerates to zero velocity in 0.9 seconds upon hitting the water, is the module structurally compromised?

$g=10\frac{m}{s^2}$

Possible Answers:

No; the impulse experienced is $4000\frac{kg\cdot m}{s}$

Yes; the impulse experienced is $5300\frac{kg\cdot m}{s}$

Yes; the impulse experienced is $4700\frac{kg\cdot m}{s}$

No; the impulse experienced is $3400\frac{kg\cdot m}{s}$

More information is needed to solve

Correct answer:

No; the impulse experienced is $4000\frac{kg\cdot m}{s}$

Explanation:

We need to use the equation for impulse to solve this problem. In fact, the time given is completely irrevelevant:

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

Plugging in our values:

$I = (500kg)(8\frac{m}{s}) = 4000 \frac{kg\cdot m}{s}$

This is less than the threshold, so no, nothing becomes structurally compromised

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Example Question #35 : Impulse And Momentum

An arrow with mass m = 0.2kg is shot straight up in the air with an initial kinetic energy of 10J . What is the momentum of the arrow after . Neglect air resistance.

$g = 10\frac{m}{s^2}$

Possible Answers:

$0\frac{kg\cdot m}{s}$

$4\frac{kg\cdot m}{s}$

$0.2\frac{kg\cdot m}{s}$

$2\frac{kg\cdot m}{s}$

$08\frac{kg\cdot m}{s}$

Correct answer:

$0\frac{kg\cdot m}{s}$

Explanation:

In order to determine the momentum of the arrow, we are going to need its velocity at that time. First, we will convert the initial kinetic energy into initial velocity:

$KE_i = \frac{1}{2}mv_i^2$

Rearranging for velocity:

$v_i = \sqrt{\frac{2KE_i}{m}}$

$v_i = \sqrt{\frac{2(10J)}{0.2kg}}$

$v_i = 10\frac{m}{s}$

Then we can use the following equation to calculate the velocity after 1 second:

v_f = v_i+at

$v_f = 10\frac{m}{s}+\left ( -10\frac{m}{s^2}\right )(1s)$

$v_f = 0\frac{m}{s}$

Then using the expression for momentum:

p = mv

$p = \left ( 0\frac{m}{s}\right )(0.2kg)$

$p= 0 \frac{kg\cdot m}{s}$

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Example Question #491 : Ap Physics 1

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

Possible Answers:

Momentum is conserved only in one-dimensional collisions.

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

Momentum is always conserved.

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

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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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #30 : Impulse And Momentum

Example Question #491 : Ap Physics 1

Example Question #31 : Impulse And Momentum

Example Question #32 : Impulse And Momentum

Example Question #31 : Impulse And Momentum

Example Question #35 : Impulse And Momentum

Example Question #491 : Ap Physics 1

Example Question #33 : Impulse And Momentum

Example Question #34 : Impulse And Momentum

Example Question #37 : Impulse And Momentum

All AP Physics 1 Resources

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