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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #64 : Impulse And Momentum

A 120kg man is running at $2.1\frac{m}{s}$ . From that run, he jumps on a resting skateboard of mass of 4.5kg . Assuming no energy lost to friction, determine the final velocity of the man on the board.

Possible Answers:

$7.88\frac{m}{s}$

$4.40\frac{m}{s}$

None of these

$1.11\frac{m}{s}$

$2.02\frac{m}{s}$

Correct answer:

$2.02\frac{m}{s}$

Explanation:

Using conservation of momentum:

p_i=p_f

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

The man and the skateboard are stuck together, and thus become one mass:

$m_1v_{1i}+m_2v_{2i}=(m_1+m_2)v_{f}$

Plugging in values:

(120)(2.1)+(4.5)(0)=(124.5)(v_f)

Solving for v_f

$v_f=2.02\frac{m}{s}$

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Example Question #71 : Impulse And Momentum

A blob in space of mass 100kg is moving in a straight line at $10\frac{m}{s}$ . It then collides and sticks to another blob that was previously stationary. The final velocity of the combined blobs is $2.5\frac{m}{s}$ . Find the mass of the second blob.

Possible Answers:

1200kg

300kg

100kg

None of these

400kg

Correct answer:

300kg

Explanation:

Using

$m_1v_1+m_2v_2=m_{1+2}v_{1+2}$

Plugging in values:

(100)(10)+(m_2)(0)=(100+m_2)(2.5)

m_2=300kg

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Example Question #72 : Impulse And Momentum

A 2.7g ping pong ball is accelerating at $<4,1>\frac{m}{s}$ . Determine the net force on the ball.

Possible Answers:

<.0108,.0027>N

<8.1,1.5>N

<9.1,2.2>N

<1.08,2.7>N

<3.18,4.7>N

Correct answer:

<.0108,.0027>N

Explanation:

Using

$\vec{F}=m\vec{a}$

Converting to and plugging in values

$\vec{F}=.0027<4,1>$

$\vec{F}=<.0108,.0027>N$

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Example Question #73 : Impulse And Momentum

A bullet weighing 60g is fired at a velocity of $360\frac{m}{s}$ at a block weighing 1kg at rest on a frictionless surface. When the bullet hits the block, it becomes lodged in the block and causes the block to move. How fast does the block move after the collision?

Possible Answers:

$6.0\frac{m}{s}$

$20.4 \frac{m}{s}$

$354 \frac{m}{s}$

$4.52\frac{m}{s}$

$21.6\frac{m}{s}$

Correct answer:

$20.4 \frac{m}{s}$

Explanation:

Using the equation of momentum,

p=mv

we can compare the initial and final scenarios and set them equal to each other to solve for the final velocity of the block (don't forget to convert the units of to !):

$m_{1}v_{1}=m_{2}v_{2}$

$.06\times360 = (.06+1)\times v_{2}$

$v_2\approx 20.4$

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Example Question #251 : Linear Motion And Momentum

A .1kg tennis ball hits a wall head-on with a forward speed of $20\frac{m}{s}$ . It rebounds with a speed of $15\frac{m}{s}$ . The contact time is .10 s . What is the impulse with the wall?

Possible Answers:

$-0.5 N \cdot s$

$3.5 N \cdot s$

$0 N \cdot s$

$0.5 N \cdot s$

$-3.5 N\cdot s$

Correct answer:

$-3.5 N\cdot s$

Explanation:

Impulse is defined as a change in momentum:

$I = m\Delta v=m(v_{f}-v_{i})=0.1(-15-20)=-3.5 N\cdot s$

Pay attention to direction, because this changes the sign!

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Example Question #75 : Impulse And Momentum

Force 1 acts on an object for 1000s with . Force 2 acts on an object for 0.1s with 10000N .

Which force changes the momentum of the object more?

Possible Answers:

Force 1

It's impossible to tell.

There is no momentum change in either situation.

Force 1 and Force 2 change the momentum of the object the same amount.

Force 2

Correct answer:

Force 1 and Force 2 change the momentum of the object the same amount.

Explanation:

The impulse momentum theorem states:

$\Delta p=F\Delta t$

Where $\Delta p$ is the change in momentum of an object, is the force acting on the object, and $\Delta t$ is the amount of time the force acts on the object.

The change in momentum due to Force 1 can be found by:

1000N*1s=1000Ns

And the change in momentum due to Force 2 can be found by:

10000N*0.1s=1000Ns

So both situations produce the same change in momentum.

It's important to understand that both force and time have a linear relationship with momentum, meaning that as either force and/or time increase, momentum will also increase.

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Example Question #261 : Linear Motion And Momentum

Ball 1 has a mass of 5kg a velocity of $4 \frac{m}{s}$ . Ball 2 has a mass of 2 kg and a velocity of $10 \frac{m}{s}$ .

Which ball has the most momentum?

Possible Answers:

The balls have equivalent momentum.

Ball 2

It's impossible to tell.

Neither ball has any momentum.

Ball 1

Correct answer:

The balls have equivalent momentum.

Explanation:

The equation for momentum is given as:

p=mv

Where is the momentum of an object, is the mass of the object, and is the velocity of the object.

For Ball 1 we have:

$p=5kg*4\frac{m}{s}=20\frac{kgm}{s}$

For Ball 2 we have:

$p=2kg*10\frac{m}{s}=20\frac{kgm}{s}$

So the balls have equivalent momentum.

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Example Question #71 : Impulse And Momentum

An object has a mass, M, and a velocity, V.

What happens to the objects momentum if its velocity is doubled?

Possible Answers:

The objects momentum is halved.

The objects momentum remains the same.

The objects momentum goes to zero.

The objects momentum is quadrupled.

The objects momentum doubles.

Correct answer:

The objects momentum doubles.

Explanation:

The equation for momentum is given as:

p=mv

Where is the momentum of an object, is the mass of the object, and is the velocity of the object.

The relationship between momentum and velocity is linear, so when the objects velocity is doubled, its momentum is also doubled.

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Example Question #261 : Linear Motion And Momentum

A space shuttle of mass 2.0Mkg is traveling through space at a velocity of $700 \frac{m}{s}$ and has gotten slightly off course. What was the impulse if the shuttles velocity is adjusted to $770\frac{m}{s}$ ?

Possible Answers:

$80MN\cdot s$

$120kN\cdot s$

$600kN\cdot s$

$140MN\cdot s$

$940N\cdot s$

Correct answer:

$140MN\cdot s$

Explanation:

We can use the expression for impulse for this problem:

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

Since we are given the mass and can easily calculate the change in velocity, we'll go with the second form of this expression:

$I = (2000000kg)\left ( 770\frac{m}{s}-700\frac{m}{s}\right )$

$I = 140MN\cdot s$

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Example Question #261 : Linear Motion And Momentum

A golfer putts a 0.046kg ball 20m away. The ball rolls this distance, never leaving the ground. If the 1kg club was swung at $1 \frac{rad}{s}$ , and the coefficient of friction between grass and ball is 0.9 , how long was the golf club and arm?

Possible Answers:

0.86m

0.95m

0.53m

1.23m

0.49m

Correct answer:

0.86m

Explanation:

Conservation of momentum states that the angular momentum of the golf swing transfers into the linear momentum of the golf ball, or $L = m_{1}\omega r = p = m_{2}v$ . Since the only forces acting on the ball are kinetic friction and gravity, we can use kinematics to put this equation completely into known quantities.

Solving for $v_{0}$ gives $v_0 = \frac{m_1 \omega r}{m_2}$ . Kinematics gives that $v^2 = v_{0}^2 + 2a\Delta x$ . Substituting the first expression into the second gives that $v^2= (\frac{m_1 \omega r}{m_2})^2+2a\Delta x$ .

The kinetic friction force can be expressed as $f_{k} = -\mu_{k}N = -\mu_{k}mg$ and by Newton's second law, $-\mu_{k}mg = ma$ . Hence the acceleration due to friction can be written as $a = -\mu_{k}g$ . Substituting this into our previous kinematic expression finally gives

$v^2 = (\frac{m_1 \omega r}{m_2})^2-2\mu_{k}g\Delta x$ . Plugging in known values gives $(\frac{(1kg)(1\frac{rad}{s})r}{(0.046kg)})^2 = 2(0.9)(9.8 \frac{m}{s^2})(20m)$ . Solving for gives r = 0.86m

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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 #64 : Impulse And Momentum

Example Question #71 : Impulse And Momentum

Example Question #72 : Impulse And Momentum

Example Question #73 : Impulse And Momentum

Example Question #251 : Linear Motion And Momentum

Example Question #75 : Impulse And Momentum

Example Question #261 : Linear Motion And Momentum

Example Question #71 : Impulse And Momentum

Example Question #261 : Linear Motion And Momentum

Example Question #261 : Linear Motion And Momentum

All AP Physics 1 Resources

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