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AP Physics 1 : Impulse and Momentum

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Example Question #51 : Impulse And Momentum

An arrow of mass 0.1kg is shot directly upward with an initial velocity of $v_i = 60\frac{m}{s}$ . What is the momentum of the arrow after and in what direction? Neglect air resistance.

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

Possible Answers:

$5\frac{kg\cdot m}{s},upward$

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

$5\frac{kg\cdot m}{s},downward$

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

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

Correct answer:

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

Explanation:

Since the arrow is shot directly upward, we can easily calculate the velocity of the arrow after 8 seconds:

v_f = v_i+at

Plugging in our values, we get:

$v_f = 60\frac{m}{s}+\left ( -10\frac{m}{s^2}\right )(8s)$

$v_f = -20\frac{m}{s}$

Which is also:

$v_f = 20\frac{m}{s}, downward$

Then using the expression for momentum, we get:

p=mv_f

$p=(1kg)\left ( 20\frac{m}{s}\right )$

$p = 2\frac{kg\cdot m}{s}, downward$

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Example Question #52 : 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.

How should the final momentum relate to the initial momentum?

Possible Answers:

There is not enough information to determine the final momentum

The final momentum will be less because the objects stuck together

The final momentum will be greater because the objects stuck together

They will be the same

Correct answer:

They will be the same

Explanation:

Momentum is always conserved in a closed system. Thus, since this system has no external forces acting on it (such as gravity, friction, electromagnetism, etc.) then momentum will be the same at the end as in the beginning.

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Example Question #51 : Impulse And Momentum

Deep in space, mass is traveling at 13.4mph , it then collides with mass . After the collision, mass is motionless. Mass then collides with and sticks to mass . Determine the final velocity of the mass mass combination.

All three of the masses are identical.

Possible Answers:

13.4mph

3.35mph

None of these

6.7mph

26.8mph

Correct answer:

6.7mph

Explanation:

The initial collision in completely elastic, all of the momentum and energy is transferred to the second mass.

m_A*13.4=m_B*v_B

The second collision is completely inelastic, all of the momentum is retained by the two objects that stick together.

$m_Bv_B=m_{B+C}v_{B+C}$

Since all three of the masses are identical:

$m_{B+C}=m_B+m_C$

m_A=m_B=m_C

Combining equations

$m_A*13.4=2m_A*v_{B+C}$

$v_{B+C}=6.7mph$

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Example Question #54 : Impulse And Momentum

A spaceship of mass 7500kg is motionless in space. The rocket is turned on and provides a constant force of 1000N . Assume the mass of mass due to spent fuel is negligible.

Determine the velocity of the ship after 10s

Possible Answers:

$35\frac{m}{s}$

$0\frac{m}{s}$

$1000\frac{m}{s}$

$2.3\frac{m}{s}$

$1.33\frac{m}{s}$

Correct answer:

$1.33\frac{m}{s}$

Explanation:

F=ma

Plugging in values

1000=7500*a

$a=.133\frac{m}{s}$

Using

v_i+a*t=v_f

0+.133*10=v_f

$v_f=1.33\frac{m}{s}$

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Example Question #51 : Impulse And Momentum

Two football players are pushing on each other during a play. The offensive player is pushing the defensive player backwards. What can be said about the forces in this situation?

Possible Answers:

The players are exerting unequal forces on each other, the offensive player is exerting more force on the defensive player.

None of these

It is impossible to determine anything without knowing the mass of each player.

The players are exerting unequal forces on each other because the offensive player is exerting more force on the ground that the defensive player.

The players are exerting equal forces on each other, however, the offensive player is exerting more force on the ground that the defensive player.

Correct answer:

The players are exerting equal forces on each other, however, the offensive player is exerting more force on the ground that the defensive player.

Explanation:

Based on Newtons laws, the players are exerting equal forces on each other. The reason the offensive player is "winning" is that he is able to push harder against the ground than the defensive player.

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Example Question #56 : Impulse And Momentum

A spaceship of mass 7500kg is motionless in space. The rocket is turned on and provides a constant force of 1000N . Assume the loss of mass due to spent fuel is negligible.

Determine the acceleration of the ship.

Possible Answers:

$.075\frac{m}{s}$

$.266\frac{m}{s}$

None of these

$.133\frac{m}{s}$

$1000\frac{m}{s}$

Correct answer:

$.133\frac{m}{s}$

Explanation:

F=ma

Plugging in values

1000=7500*a

$a=.133\frac{m}{s}$

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Example Question #51 : Impulse And Momentum

A spaceship of mass 7500kg is motionless in space. The rocket is turned on and provides a constant force of 1000N . Assume the loss of mass due to spent fuel is negligible.

Determine the momentum of the ship after 10s .

Possible Answers:

None of these

$4875kg\cdot \frac{m}{s}$

$1375kg\cdot \frac{m}{s}$

$2575kg\cdot \frac{m}{s}$

$9975kg\cdot \frac{m}{s}$

Correct answer:

$9975kg\cdot \frac{m}{s}$

Explanation:

F=ma

Plugging in values

1000=7500*a

$a=.133\frac{m}{s}$

Using

v_i+a*t=v_f

0+.133*10=v_f

$v_f=1.33\frac{m}{s}$

Using

p=mv

7500*1.33=p

$p=9975kg\cdot \frac{m}{s}$

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Example Question #51 : Impulse And Momentum

Tom drops a ball of mass from rest from a height . The ball bounces back to a height of $\frac{h}{2}$ . Find the magnitude of the impulse the ground imparted on the ball.

Possible Answers:

$m\sqrt{gh}$

$(\sqrt{2}-1)m\sqrt{gh}$

$m\sqrt{2gh}$

$2m\sqrt{2gh}$

$(1+\sqrt{2})m\sqrt{gh}$

Correct answer:

$(1+\sqrt{2})m\sqrt{gh}$

Explanation:

Impulse is just the change in momentum. To find the velocity when the ball hits the ground, we need to use kinematics. We know the height the ball is dropped, the acceleration, and the initial velocity, so we can use the equation $v_{final}^2=v_{initial}^2+2a\Delta x$ . The initial velocity is , a=g , and $\Delta x=h$ , so the equation becomes $v_{final}^2=2gh$

$v_{final}=\sqrt{2gh}$

When the ball bounces back up it reaches a height of $\frac{h}{2}$ . In order to find the velocity immediately after it hits the ground, we can use the same equation with $\Delta x=\frac{h}{2}$ . This will lead it a velocity of $\sqrt{2g*\frac{h}{2}}=\sqrt{gh}$

Assuming up is positive, the magnitude of the impulse is just $m\left | \sqrt{gh}-(-\sqrt{2gh})\right |=(1+\sqrt{2})m\sqrt{gh}$

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Example Question #51 : Impulse And Momentum

Two identical rockets are placed on the back of cars. Car has a mass twice that of car . The rockets are identically fired for 10s , then shut off.

How will the momentum of each car compare?

Possible Answers:

None of these

The heavier car has more momentum

The lighter car has more momentum

Impossible to determine

They are the same

Correct answer:

They are the same

Explanation:

Using

$J=F*\Delta t$

Where represents the impulse, which is the change in momentum.

It can be seen that since both cars have the same force applied for equal amount of time, they have identical final momentums.

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Example Question #52 : Impulse And Momentum

An object's momentum as a function of time is the following:

p(t)=3k^3 , where is some constant.

What is the force of the object (as a function of time) that causes the motion?

Possible Answers:

3k^2

9k^2

k^4

Correct answer:

9k^2

Explanation:

$\vec{F}=\frac{d\vec{p}}{dt}.$

Therefore, we simply need to differentiate our momentum equation with respect to time to determine the force.

$\vec{F}=\frac{d\vec{p}}{dt}=\frac{d(3k^3)}{dt}=9k^2$

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AP Physics 1 : Impulse and Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #51 : Impulse And Momentum

Example Question #52 : Impulse And Momentum

Example Question #51 : Impulse And Momentum

Example Question #54 : Impulse And Momentum

Example Question #51 : Impulse And Momentum

Example Question #56 : Impulse And Momentum

Example Question #51 : Impulse And Momentum

Example Question #51 : Impulse And Momentum

Example Question #51 : Impulse And Momentum

Example Question #52 : Impulse And Momentum

All AP Physics 1 Resources

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