Motion in One Dimension

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AP Physics 1 › Motion in One Dimension

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A hungry wasp spots an fly wandering about. Assuming the wasp attacks the fly from behind (they are both traveling in the same direction) with speed v, and the fly is stationary, what is the speed of the wasp and fly after the collision? Assume the fly and wasp are one object after the collision. Your answer should be in terms of M, m, v where M is the mass of the wasp, m is the mass of the fly and v is the original speed of the wasp.

$\frac{Mv}{M+m}$

CORRECT

$\frac{mv}{M}$

$\frac{M+m}{Mv}$

$\frac{mv}{M+m}$

, they are both stationary after the collision.

Explanation

Considering the wasp aims to eat the fly, we assume the fly and wasp are one body after the collision. This is an inelastic collision. We can solve this with conservation of momentum.

$\Delta P=0$ or $P_{f}=P_{i}$

For the two body inelastic colision between the wasp and the fly, we can rewrite this as:

$(mv)_{wasp}+(mv)_{fly}=(m_{wasp}+m_{fly})v_{fly+wasp}$

Then taking into account the fact the fly is stationary initially:

$Mv+0=(M+m)v_{fly+wasp}$

Then solve for the velocity of the fly and the wasp after the collision:

$v_{fly+wasp}=\frac{Mv}{M+m}$

Consider a particle initially located at and moving with initial velocity $v_0 = -3 \frac{m}{s}$ . Assuming a constant acceleration of $a=5 \frac{m}{s^2}$ , calculate the position at a time of .

CORRECT

Explanation

Looking at the initial information we are given about the particle at , we can construct the equation of motion for the position of the particle as:

$x(t)=\frac{1}{2}at^2+v_0t+x_0$

Plug in our values and solve.

$x(6)=\frac{1}{2}(5\frac{m}{s^2})(6s)^2-3\frac{m}{s}(6s)+2m=74m$

Suppose that a ball is thrown straight upward and falls back to the ground in a time . If this same ball is thrown straight upward on a distant planet whose gravity is only one-third that of Earth's, then will change by what factor?

Increase by a factor of

CORRECT

Decrease by a factor of

Increase by a factor of $\sqrt{3}$

Decrease by a factor of $\sqrt{3}$

Explanation

In this question, we're being asked to determine how long a ball will remain in the air when it is thrown vertically upward on a planet with reduced gravity. First, we'll need to find an expression that relates gravity with the amount of time the ball remains in the air. To do this, we can make use of the kinematics equations.

$\Delta y=v_{iy}t+\frac{1}{2}at^{2}$

Furthermore, since we know the ball will land where it began, we know that $\Delta y=0$ .

$\Delta y=0=v_{iy}t+\frac{1}{2}at^{2}$

$v_{iy}t=-\frac{1}{2}at^{2}$

$v_{iy}=-\frac{1}{2}at$

Moreover, if we define the upward direction as positive and the downward direction as negative, then we know that , since gravity is always pointing in the downward direction.

$v_{iy}=\frac{1}{2}gt$

$t=\frac{2v_{yi}}{g}$

The above expression is the one we're looking for because it relates time and gravity. From this expression, we can conclude that if the magnitude of gravity is reduced by a factor of three, then the time variable will increase by a factor of three.

Boomervt1

According to the graph shown above, during which interval is Boomer at rest?

$7.5s\rightarrow 10s$

CORRECT

$0s\rightarrow2.5s$

$2.5s\rightarrow5s$

5 - 7.5 s

None of these

Explanation

Boomer is at rest during periods when the slope of the position versus time graph is zero.

A person travelling at a rate of $36\frac{m}{s}$ , with initial position at will have travelled to in how much time?

CORRECT

Explanation

This is a simple question of rate, time, and distance.

$d=r\cdot t$ , where is distance travelled, is rate, is time passed.

In our case, we know the rate is

$r=36 \frac{m}{s}$

We also know that the person travelled to having originally started at at . Keeping this in mind, the distance travelled is:

Now we just solve for time:

$t=\frac{d}{r}=\frac{540m}{36\frac{m}{s}}=15s$

My roommate locked himself out of the apartment and forgot his keys. He asks me to walk over to the balcony and drop my key off the edge so he can get in. The keys take to reach the ground from the time it is released from my hand. From what height were the keys released, neglect the effects due to air resistance?

CORRECT

Explanation

Use a kinematic equation to solve for the height of the building.

$X_f=X_i+V_o(\Delta t)+\frac{1}{2}(a)(\Delta t)^{2}$

Plug in and solve for the height of the building.

$0m=X_i+0\frac{m}{s}(6.0s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}(6.0s)^{2}$

$0m=X_i+(-4.9\frac{m}{s^{2}})(36s^{2})$

The building was tall.

A wreak-less card driver was driving in the eastward direction at $65\frac{m}{s}$ when he noticed that the car in front of him was at a complete halt. He subsequently slammed the brakes causing an acceleration of $-7.5\frac{m}{s^{2}}$ . Will he be saved by his brakes or will he hit the car that was in front of him when he first applied the brakes?

Would hit the car; or stop at

CORRECT

Stops safely at

Stops just safely at

Hits the car; or would stop at

Explanation

Use a kinematic equation to solve for the total time.

$V_f=V_o+a\Delta t$

Plug in and solve for time.

$0\frac{m}{s}=65\frac{m}{s}+(-7.5\frac{m}{s^{2}})(t_{f}-0s)$

$-65\frac{m}{s}=(-7.5\frac{m}{s^{2}})(t_{f})$

$t_{f}=8.66 s$

Use another kinematic equation to solve for the final distance.

$X_f=X_i+V_o(\Delta t)+\frac{1}{2}(a)(\Delta t)^{^{2}}$

Plug in the known values and solve for the final distance.

$X_{f}=0m+65\tfrac{m}{s}(8.66s)+\frac{1}{2}(-7.5\frac{m}{s}^{2})(8.66s)^{2}$

$X_{f}=563.3m+(-32.47m)$

$X_{f}=530.8 m$

The car would have stopped at , however the other car was at thus there would have unfortunately been a car accident.

If a 15kg ball takes five seconds to strike the ground when released from rest, at what height was the ball dropped?

125m

CORRECT

100m

50m

250m

75m

Explanation

Using the equation we can find the distance at which the ball was dropped. Notice that the mass of the ball does not matter in this problem. We are told that the ball is dropped from rest making, , thus we have . When we plug in our values, and assuming that acceleration is equal to gravity (10m/s2) we find that = 125m.

Boomerstablevt

According to the table above, during which interval does Boomer reach his highest speed?

None: there is a tie

CORRECT

$0s\rightarrow2.5s$

$2.5s\rightarrow5s$

$5s\rightarrow7.5s$

$7.5s\rightarrow10s$

Explanation

Boomer's speed is the ratio of the distance traveled to the time. It is $2\frac{m}{s}$ during both the first and and third intervals.

A car traveling at $100: \frac{m}{s}$ suddenly applies the brakes until it comes to a stop. If the car decelerates at a constant rate of $5: \frac{m}{s^{2}}$ , how long will it take the car to come to a stop?

CORRECT

Explanation

To answer this question, we'll first need to find the distance the car travels before coming to a stop. Since we are told in the question stem that the car is decelerating at a constant rate, we know that we have a situation in which acceleration is constant and thus we can make use of the kinematic equations.

$v_{f}^{2}=v_{i}^{2}+2a\Delta x$

$\Delta x=\frac{v_{f}^{2}-v_{i}^{2}}{2a}$

And since we know that the car is coming to a stop, we know that our final velocity will be equal to zero.

$\Delta x=\frac{0-v_{i}^{2}}{2a}$

Furthermore, since we know the car is decelerating, we know it is slowing down and is thus accelerating in the direction opposite to its direction of motion. So if we assign a positive value to the velocity, then the acceleration will have a negative value.

$\Delta x=\frac{-v_{i}^{2}}{-2a}=\frac{v_{i}^{2}}{2a}$

$\Delta x=\frac{(100 \frac{m}{s})^{2}}{2(5 \frac{m}{s^{2}})}=\frac{10000 \frac{m^{2}}{s^{2}}}{10 \frac{m}{s^{2}}}=1000m$

Now that we have found the distance traveled by the car, we can calculate the amount of time it takes for the car to come to a stop.

$\Delta x=v_{i}t+\frac{1}{2}at^{2}$

Plug in known values and solve.

$1000 m=100 \frac{m}{s}t+\frac{1}{2}(-5\frac{m}{s^{2}})t^{2}=100 \frac{m}{s}(t)- \frac{5m}{2s^{2}}(t^{2})$

From the above expression, we see that there are two solutions that can satisfy the value of . Only one of them, however, will make any sense physically. To solve for the two values, we will need to use the quadratic equation.

$-\frac{5m}{2s^2}(t^{2})+100 \frac{m}{s}(t)-1000m=0$

$t=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

Our known values for the variables above are as follows:

$a=-\frac{5}{2}$

Plug into the quadratic equation and solve.

and . The negative value makes no sense physically, thus our answer is .

An alternative (and much faster) way of solving this problem that yields the same answer is to use the following equation.

$v_{f}=v_{i}+a_{average}t$

We know that our final velocity is equal to zero since the car is coming to a stop, and we're told in the question stem that we're dealing with a constant deceleration (negative acceleration).

$0=100 \frac{m}{s}+-5\frac{m}{s^{2}}t$

$-100\frac{m}{s}=-5\frac{m}{s^{2}}t$

$t=\frac{100 \frac{m}{s}}{5 \frac{m}{s^{2}}}=20s$

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