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AP Physics 1 : Linear Motion and Momentum

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Example Questions

Example Question #61 : Motion In One Dimension

A car traveling at $20 \frac{m}{s}$ sees a red light ahead and begins to slow down. The gas pedal is not touched until after the car has left the intersection. It takes to reach the light, which turns green again before the car comes to a complete stop. At this point the car rolls through the intersection at $5 \frac{m}{s}$ . What was the car's acceleration during this time?

Possible Answers:

$-.33 \frac{m}{s^{2}}$

$3 \frac{m}{s^{2}}$

$.33 \frac{m}{s^{2}}$

$-3 \frac{m}{s^{2}}$

Correct answer:

$-3 \frac{m}{s^{2}}$

Explanation:

The equation for acceleration is $\frac{\Delta v}{\Delta t}$ . The final velocity minus the initial velocity is $-15 \frac{m}{s}$ and the change in time is . When dividing the change in velocity by the change in time you receive the answer $-3 \frac{m}{s^{2}}$ which means that the car is decelerating due to the negative sign.

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Example Question #291 : Newtonian Mechanics

While riding a bike down a steep hill at $9\frac{m}{s}$ you skid to a stop in 3.2 seconds. How far did the bike slide? Assume uniform acceleration.

Possible Answers:

$\Delta x=12.0m$

$\Delta x=4.4m$

$\Delta x=14.4m$

$\Delta x=1.4m$

$\Delta x=10.4m$

Correct answer:

$\Delta x=14.4m$

Explanation:

$\Delta x= (\frac{V_{i}+V_{f}}{2})*t=\frac{9\frac{m}{s}}{2}*3.2s=\mathbf{14.4m}$

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Example Question #332 : Ap Physics 1

A motorcycle accelerated from rest to $12\frac{m}{s}$ over a distance of 50 meters. What is the rider's acceleration?

Possible Answers:

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

$a=1.44\frac{m}{s^2}$

None of these.

$a=.744\frac{m}{s^2}$

$a=2.2\frac{m}{s^2}$

Correct answer:

$a=1.44\frac{m}{s^2}$

Explanation:

To find the acceleration, use the kinematic equation $v^2=V_{0}^2+2a(x-x_{0})$

$V_{f}=12\frac{m}{s}$

$V_{0}=0$

$(12\frac{m}{s})^2=2a(50m)$

$a=\frac{144\frac{m^2}{s^2}}{100m}=\boldsymbol{1.44\frac{m}{s^2}}$

*the rider's acceleration will match that of the motorcycle.

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Example Question #64 : Linear Motion And Momentum

A quarterback throws a football a horizontal distance of 50 yd to a wide receiver. The ball was airborne for 5 s . The ball had an initial speed of 50 mph . The ball has a mass of .43kg .

$1mph=.447\frac{m}{s}$

1 yd=.914 m

Determine the maximum height of the football.

Possible Answers:

16.81m

23.98m

18.25m

None of these

21.23m

Correct answer:

21.23m

Explanation:

Determining horizontal component of velocity:

d_x=v_x*t

$v_x=\frac{d_x}{t}$

$v_x=\frac{45.7}{5}$

$v_x=9.14\frac{m}{s}$

Using

v_x^2+v_y^2=v^2

Solving for v_y

$v_y=\sqrt{v^2-v_x^2}$

Combining equations

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

Converting mph to mps

$50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}$

Converting yards to meters

$50yards*\frac{.914 meters}{1 yard}=45.7 meters$

Plugging in values:

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

$v_y=\sqrt{22.35^2-(\frac{45.7}{5})^2}$

$v_y=20.4\frac{m}{s}$

The ball will be at it's maximum height when the component of it's velocity, and thus that direction of kinetic energy, is zero.

The velocity will be decreased by the negative work done by gravity.

.5mv_y_i^2+mgh=.5mv_i_f^2

Plugging in values:

.5*20.4^2+(-9.8)h=.5*0^2

Solving for

h=21.23meters

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Example Question #61 : Linear Motion And Momentum

A projectile is fired at a wall with a speed of $400\frac{m}{s}$ . It penetrates the wall to a distance of 0.053 meters before stopping. What is the acceleration of the projectile within the wall? Assume uniform acceleration.

Possible Answers:

$a=-5.11*10^6\frac{m}{s^2}$

$a=-2.51*10^6\frac{m}{s^2}$

$a=-1.51*10^6\frac{m}{s^2}$

$a=5.11*10^6\frac{m}{s^2}$

$a=1.51*10^6\frac{m}{s^2}$

Correct answer:

$a=-1.51*10^6\frac{m}{s^2}$

Explanation:

Use the kinematic equation $V_{f}^2=V_{i}^2+2a(x-x_{0})$ .

$0=(400\frac{m}{s})^2+2a(0.053m)$

$-160,000\frac{m^2}{s^2}=a(0.106m)$

$\frac{-160,000\frac{m^2}{s^2}}{0.106m}=a$

$\mathbf{a=-1.51*10^6\frac{m}{s^2}}$

Negative acceleration because the projectile was slowing down.

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Example Question #65 : Motion In One Dimension

A motorcycle must be traveling at $12\frac{m}{s}$ before leaving the end of a platform to clear a particular jump. If the motorcycle accelerates down the length of the ramp at $2.8\frac{m}{s^2}$ for 5 seconds will it clear the jump?

Possible Answers:

No, it will be moving $2\frac{m}{s}$ slower than necessary

Yes, it will be moving $4\frac{m}{s}$ faster than necessary

No, it will be moving $4\frac{m}{s}$ slower than necessary

There is not enough information to determine

Yes, it will be moving $2\frac{m}{s}$ faster than necessary

Correct answer:

Yes, it will be moving $2\frac{m}{s}$ faster than necessary

Explanation:

To find its speed before leaving the ramp, $V_{f}=V_{0}+at\rightarrow 2.8\frac{m}{s^2}*5s=14\frac{m}{s}$ .

Yes, it will be moving $2\frac{m}{s}$ faster than necessary.

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Example Question #62 : Linear Motion And Momentum

You discover an old well in the forest behind your house. In order to ascertain its depth you drop a rock which falls straight down to the surface of the water. How deep is the well if it takes 5 seconds for the rock to hit the water (neglecting air resistance and the time it took for the sound to travel)?

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

Possible Answers:

$225\hspace{1mm}m$

$125\hspace{1mm}m$

$75\hspace{1mm}m$

$154\hspace{1mm}m$

None of these

Correct answer:

$125\hspace{1mm}m$

Explanation:

Use the kinematic equation $\Delta x= x_{0}+v_{0}t+.5at^2$ .

Since the object was dropped and starts at position 0 in the hand, the first two terms initial position and initial velocity become 0.

$\Delta x=.5at^2 \rightarrow x=.5(10\frac{m}{s^2})(5)^2\rightarrow \boldsymbol{125 m}$

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Example Question #63 : Linear Motion And Momentum

A theme park ride raises the participants to a bird's eye view and then drops them. If it take 3 seconds for the ride to reach its stopping position below, what was the final speed before coming to a halt just above the ground? Ignore frictional forces.

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

Possible Answers:

$V_{f}=6.2 \frac{m}{s}$

$V_{f}=31 \frac{m}{s}$

$V_{f}=62 \frac{m}{s}$

$V_{f}=3.11 \frac{m}{s}$

$V_{f}=311\frac{m}{s}$

Correct answer:

$V_{f}=31 \frac{m}{s}$

Explanation:

Since the object is just falling it is only subject to the acceleration of gravity (especially since we are ignoring frictional forces).

So the final speed is equal to the time multiplied by the acceleration of gravity: $V_{f}=10 \frac{m}{s^2}*3.1 s= \boldsymbol{31 \frac{m}{s}}$

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Example Question #68 : Motion In One Dimension

A man stands on a building 10m above the ground. He reaches over the side and throws an apple straight up with at a speed of $20\frac{m}{s}$ .

How long does it take for the apple to hit the ground below?

Possible Answers:

4.53s

5.25s

3.75s

1.75s

4.08s

Correct answer:

4.53s

Explanation:

The apple is flying straight up with an initial velocity of $20\frac{m}{s}$ , with the only force acting on it being gravity. Gravity will provide the apple with an acceleration of $9.8\frac{m}{s^2}$ directed downwards. The following kinematic equation is needed to solve for the time it takes for the apple to hit the ground:

v_f=v_i+a*t

The first step is to see how long it takes the apple to get to the highest point of its journey. At that point, the velocity will have slowed to zero:

$v_i=20\frac{m}{s}$

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

$a=-9.8\frac{m}{s^2}$

Solving for t gives us:

$t=\frac{v_f-v_i}{a}=\frac{0-20}{-9.8}=2.04 s$

Since it took the apple 2.04 seconds to get to the top, that means it will take the apple another 2.04 seconds to get back to its starting point. This also means that the velocity of the apple when it reaches is starting point will be equal and opposite its starting velocity:

$v_f=v_i+a*t=0-9.8(2.04)=-20\frac{m}{s}$

To find the amount of time it takes the apple to reach the ground, the following kinematic equations are used:

$v_f^2=v_i^2+2ad=-20^2+2*-9.8*-10=596\frac{m^2}{s^2}$

$v_f=\sqrt596=24.41\frac{m}{s}$

And:

v_f=v_i+a*t

$t=\frac{v_i-v_f}{a}=\frac{-20+24.41}{9.8}=0.45s$

Therefore, the total time it takes for the apple to reach the ground is:

$t_{total}=t_1+t_2+t_3=2.04+2.04+0.45=4.53s$

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Example Question #69 : Motion In One Dimension

Mark, Jim, and David apply forces to an object of mass 3 kg . The object is on level ground where the coefficient of friction is . Mark applies a 40N force to the right, Jim applies a force to the left, and David applies a 3 N force down. How far will the object travel (in meters) after seconds? Assume it starts at rest and the coefficient of static friction is low enough for the object to start moving with the mentioned forces.

Possible Answers:

33m

98.67m

72m

10.67m

Correct answer:

72m

Explanation:

We can split this question up into the x-direction and y-direction. In the y-direction, the object's acceleration is zero since it's going to remain in contact with the ground. We now can write our $\sum F=ma$ equation (let the up direction be positive). Since it's not accelerating, we have

$\sum F=ma=0\rightarrow F_{n}-mg-3=0$ .

$F_{n}$ denotes the normal force. We know that the mass is 3kg and that $g\approx10\frac{m}{s^2}$ , so we can solve for the normal force, which we'll later use to find the force of friction. $F_{n}-(3kg)(10\frac{m}{s^2})-3=0$

$F_{n}=33N$

The force of friction $(F_{f})$ is equal to $\mu F_{n}$ (where $\mu$ denotes the coefficient of friction), so

$F_{f}=(.3)(33N)=9.9N$

In the x-direction, we'll denote right as positive. The forces in the x-direction are 40N right, left, and 9.9N . Now we can write our $\sum F=ma$ equation as 40N-3N-9.9N=(3kg)a

$\frac{27.1N}{3kg}=9.0\frac{m}{s^2}$

Using this acceleration, we can use kinematics to solve for the distance the object will travel in seconds. We have time and acceleration, but we need distance, so we'll use the equation: $x(t)=(v_{initial})t+0.5at^2$

$v_{initial}=0$

$x=0.5(9.0\frac{m}{s^2})(4s)^2=72m$

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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 #61 : Motion In One Dimension

Example Question #291 : Newtonian Mechanics

Example Question #332 : Ap Physics 1

Example Question #64 : Linear Motion And Momentum

Example Question #61 : Linear Motion And Momentum

Example Question #65 : Motion In One Dimension

Example Question #62 : Linear Motion And Momentum

Example Question #63 : Linear Motion And Momentum

Example Question #68 : Motion In One Dimension

Example Question #69 : Motion In One Dimension

All AP Physics 1 Resources

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