AP Physics 1 - Motion in One Dimension

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 .

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.

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

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.

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

$\frac{mv}{M}$

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

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

, 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:

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

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?

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.

Boomervt1

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

$7.5s\rightarrow 10s$

$0s\rightarrow2.5s$

$2.5s\rightarrow5s$

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?

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:

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

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

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.

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

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.

Boomerstablevt

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

None: there is a tie

$0s\rightarrow2.5s$

$2.5s\rightarrow5s$

$5s\rightarrow7.5s$

$7.5s\rightarrow10s$

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.

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

125m

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.