Tag Archives: Physics Solutions

College Physics by Openstax Chapter 2 Problem 52

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

Solution:

Consider Figure 1. The object was dropped from a height of 75.0 m. At the start of motion, the velocity is zero, $v_{oy}=0$ .

The object traveled for a period of time t for the whole 75.0 m distance to the ground.

Part A

We are solving for the distance traveled by the object for the first 1 second. So, we have

\begin{align*} \Delta y & = v_{oy} t + \frac{1}{2}at^2\\ \Delta y & = \left( 0 \ \text{m/s} \right)\left( 1 \ \text{s} \right)+\frac{1}{2}\left( -9.81 \ \text{m/s}^2 \right)\left( 1 \ \text{s} \right)^2 \\ \Delta y & = 0 -4.905 \ \text{m} \\ \Delta y & = -4.91 \ \text{m} \\ |\Delta y| & =4.91 \ \text{m} \end{align*}

The negative sign of Δy indicates that the direction of the displacement is downward. Since we are looking for the scalar value of the distance, the answer is 4.91 m.

Part B

So we now consider the two positions of the object as shown in the figure to the right. The initial height of the object is 75.0 m above the ground, and the initial velocity is 0.

At the ground, we know that the position of the object is 0 m above the ground, but we do not know the time and velocity. Therefore, to determine the velocity of the object at this point, we proceed as follows:

\begin{align*} \left( v_2 \right)^2 & =\left( v_1 \right)^2+2a \Delta y \\ \left( v_2 \right)^2 & =\left( v_1 \right)^2+2a \left( y_2-y_1 \right) \\ v_2 & = \pm \sqrt{\left( v_1 \right)^2+2a \left( y_2-y_1 \right)}\\ v_2 & = \pm \sqrt{\left( 0\ \text{m/s} \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0 \ \text{m}-75\ \text{m} \right)} \\ v_2 & = \pm \ 38.4\ \text{m/s}\\ v_2 & =- 38.4\ \text{m/s}\\ \end{align*}

Since the object is directing downwards when it hit the ground, the velocity is negative.

Part C

First, we calculate the total time of the object’s motion from the beginning to the ground.

\begin{align*} \Delta y & =\bcancel{v_{oy}t}+ \frac{1}{2}at^2 \\ \Delta y & = \frac{1}{2}at^2 \\ 0 \text{m}-75\ \text{m} & = \frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)t^2 \\ t^2& =\frac{-75\ \text{m}}{-4.905 \text{m/s}^2}\\ t&=\sqrt{\frac{75\ \text{m}}{4.905 \text{m/s}^2}}\\ t&=3.91 \ \text{s} \end{align*}

Second, determine the total distance traveled from 0 s to 2.91 s, leaving out the last 1 s of the motion.

\begin{align*} \Delta y & = v_{oy} t + \frac{1}{2}at^2\\ \Delta y & = \left( 0 \ \text{m/s} \right)\left( 1 \ \text{s} \right)+\frac{1}{2}\left( -9.81 \ \text{m/s}^2 \right)\left( 2.91 \ \text{s} \right)^2 \\ \Delta y & = 0 -41.5 \ \text{m} \\ \Delta y & = -41.5 \ \text{m} \\ |\Delta y| & =41.5 \ \text{m} \end{align*}

Finally, subtract this distance from the total distance traveled to get the distance traveled in the last 1 second.

\begin{align*} y_{_{\text{last 1 sec}}} & = 75.0 \ \text{m}-41.5 \ \text{m} \\ y_{_{\text{last 1 sec}}} & = 33.5\ \text{m} \end{align*}

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Skidding on a Curve: Unbanked Curves – Uniform Circular Motion Example Problem

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is $\mu _s=0.60$ ; (b) the pavement is icy and $\mu _s=0.25$ .

Solution:

The forces on the car are gravity mg downward, the normal force F_N exerted upward by the road, and a horizontal friction force due to the road. They are shown in the free-body diagram of the car below. The car will follow the curve if the maximum static friction force is greater than the mass times the centripetal acceleration.

Part A

In the vertical direction (y) there is no acceleration. Newton’s second law tells us that the normal force on the car is equal to the weight mg since the road is flat:

\begin{align*} \sum_{}^{}F_y & =m\ a_c\\ \\ F_N\ -\ mg & =0\\ \\ F_N&=mg\\ \\ F_N &=\left( 1000\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)\\ \\ F_N &=9810\ \text{N} \end{align*}

In the horizontal direction the only force is friction, and we must compare it to the force needed to produce the centripetal acceleration to see if it is sufficient. The net horizontal force required to keep the car moving in a circle around the curve is

\begin{align*} \sum_{}^{}F_c & =m\ a_c\\ \\ & =m\cdot \frac{v^2}{r}\\ \\ & =\left( 1000\ \text{kg} \right)\cdot \frac{\left( 15\ \text{m/s} \right)^2}{50\ \text{m}}\\ \\ &=4500\ \text{N} \end{align*}

Now we compute the maximum total static friction force (the sum of the friction forces acting on each of the four tires) to see if it can be large enough to provide a safe centripetal acceleration. For (a), $\mu _s=0.60$ , and the maximum friction force attainable is

\begin{align*} \sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\ \\ &=\left( 0.60 \right)\left( 9810\ \text{N} \right)\\ \\ &=5886\ \text{N} \end{align*}

Since a force of only 4500 N is needed, and that is, in fact, how much will be exerted by the road as a static friction force, the car can follow the curve.

Part B

The maximum static friction force possible is

\begin{align*} \sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\ \\ &=\left( 0.25 \right)\left( 9810\ \text{N} \right)\\ \\ &=2452.5\ \text{N} \end{align*}

The car will skid because the ground cannot exert sufficient force (4500 N is needed) to keep it moving in a curve of radius 50 m at a speed of 54 km/h.

Acceleration of a Revolving Ball – Uniform Circular Motion Example

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Figure 1: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.

Solution:

The linear velocity of the ball can be computed by dividing the total arc length traveled by the total time of travel. That is, the ball traveled 2 revolutions (twice the circumference of the circle) for 1 second. Thus,

\begin{align*} \text{v} &= \frac{2\cdot2 \pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi\left( 0.600\ \text{m} \right)}{1 \ \text{s}} \\ \\ & = 7.54 \ \text{m/s} \end{align*}

Since the linear velocity has already been computed, we can now compute for the centripetal acceleration, a_c.

\begin{align*} \text{a}_\text{c} & = \frac{\text{v}^{2}}{\text{r}} \\ \\ & = \frac{\left( 7.54\ \text{m/s} \right)^{2}}{0.600\ \text{m}}\\ \\ & =94.8 \ \text{m/s}^{2} \end{align*}

College Physics 2.51 – Time of the hiker to move out from a falling rock

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution:

Part A

We know that the initial height, $y_0$ of the rock is 105 meters, and the initial velocity, $v_0$ is zero. We shall solve for the distance traveled by the rock for 1.5 seconds from the initial position first to find the height at detection.

The change in height is

\displaystyle \begin{aligned} \Delta \text{y}&=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2 \\ &=\left( 0 \right)\left( 1.50 \ \text{s} \right)+\frac{1}{2}\left( 9.81\ \text{m/s}^{2} \right)\left( 1.50\ \text{s} \right)^{2}\\ &=0+11.036\ \text{m} \\ &=11.04 \ \text{m} \end{aligned}

So, the rock falls about 11.04 m from the initial height for 1.50 seconds. Therefore, the height of the rock above his head at this point is

\displaystyle \begin{aligned} \text{y}&=\text{y}_{0}-\Delta \text{y} \\ &=105\ \text{m}-11.04\ \text{m} \\ &=93.96 \ \text{m} \end{aligned}

Part B

We shall solve for the total time of travel, that is, from the initial position to his head. Then we shall subtract 1.50 s from that to solve for the unknown time of moving out. The total time of travel is

\begin{aligned} \text{y} & =\frac{1}{2}\text{at}^{2} \\ &\text {Solving for t, we have}\\ \text{t}&=\sqrt{\frac{\text{2y}}{\text{a}}} \\ &=\sqrt{\frac{2\left( 105\ \text{m} \right)}{9.81 \ \text{m/s}^{2}}} \\ &=4.63 \ \text{s} \end{aligned}

Therefore, to move out the hiker has about

\begin{aligned} \text{t}&=4.63 \ \text{s}-1.50\ \text{s}\\ &=3.13\ \text{s} \end{aligned}

College Physics 2.50 – Motion of a Jumping Kangaroo

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Part A

The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.

From

\begin{aligned} \text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\ \end{aligned}

we have

\begin{aligned} \text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay}\\ \text{v}^2-2\text{ay} &= \left ( \text{v}_0\right)^2\\ \text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \end{aligned}

Substituting the known values,

\begin{aligned} \text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \\ \text{v}_0&=\sqrt{0^2-2\left(-9.81 \text{m/s}^2\right)\left(2.50 \text{m}\right)}\\ \text{v}_0&= {\color{green}7.00 \ \text{m/s}} \end{aligned}

Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.

Part B

Since the motion of the kangaroo has uniform acceleration, we can use the formula

\text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2

The initial and final position of the kangaroo will be the same, so $y$ is equal to zero. The initial velocity is 7.00 m/s, and the acceleration is -9.81 m/s².

\begin{aligned} \text{y} & =\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2\\ 0 & = \left( 7.00\ \text{m/s} \right)\text{t}+\frac{1}{2}\left( -9.81\ \text{m/s}^{2} \right)\text{t}^2\\ 0 & =7\text{t}-4.905\text{t}^{2}\\ 7\text{t}-4.905\text{t}^{2}&=0 \\ \text{t}\left( 7-4.905\text{t} \right) & =0 \\ \text{t}=0 \qquad &\text{or} \qquad 7-4.905\text{t}=0 \\ \end{aligned}

Discard the time 0 since this refers to the beginning of motion. Therefore, we have

\begin{aligned} 7-4.905\text{t} &=0 \\ 4.905\text{t} & = 7 \\ \text{t} & =\frac{7}{4.905} \\ \text{t}&={\color{green}1.43 \ \text{s}} \end{aligned}

The kangaroo is about 1.43 seconds long in the air.

Physics

College Physics by Openstax

Physics for Scientists and Engineers: A Strategic Approach Third Edition by Randall D. Knight

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics 3rd Edition by Randall D. Knight

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College Physics by Openstax Chapter 3 Problem 37

Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The baseline is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose service line is 6.40 m from the net?

Solution:

Note: The publication of the solution to this problem is on its way. Sorry for the inconvenience.

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College Physics by Openstax Chapter 3 Problem 36

The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

Solution:

We are required to solve for the maximum distance. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:

The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
The jumper is on level ground, and the motion started from the ground.

The formula for range is

\text{R}=\frac{\text{v}_{\text{o}}^2\sin 2\theta _{\text{o}}}{\text{g}}

Since we are already given the necessary details, we can now solve for the range.

\begin{align*} \text{R}&=\frac{\left(9.5\:\text{m/s}\right)^2\:\sin 90^{\circ} }{9.81\:\text{m/s}^2}\\ \text{R}&=9.20\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

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College Physics by Openstax Chapter 3 Problem 35

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Solution:

We are required to solve for the distance in a standing broad jump. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:

The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
The jumper is on level ground.

The formula for the range is

\text{R}=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}}}{\text{g}}

To find the initial velocity of the jump, vo, we shall use the kinematic formula from the crouch position to the time the person leaves the ground.

\text{v}_{\text{f}}^2=\text{v}_{\text{o}}^2+2\text{ax}

In this case, the final velocity will be the initial velocity of the jump.

\begin{align*} \text{v}_{\text{f}}=\sqrt{\left(0\:\text{m/s}\right)^2+2\left(1.25\times 9.81\:\text{m/s}^2\right)\left(0.600\:\text{m}\right)}=3.84\:\text{m/s} \end{align*}

So, the initial velocity of the flight is 3.84 m/s. We can now use the formula for range.

\begin{align*} \text{R}&=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta_{\text{o}}}{\text{g}} \\ \text{R}&=\frac{\left(3.84\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}\\ \text{R}&=1.50\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 3 Problem 34

An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later.
(a) What is the height of the cliff?
(b) What is the maximum height reached by the arrow along its trajectory?
(c) What is the arrow’s impact speed just before hitting the cliff?

Solution:

Consider the following illustration:

An arrow shot at a height of 1.5 m towards a cliff of height H

Part A

We are required to solve for the value of H. We shall use the formula

\Delta \text{y}=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2

or, we can also write the formula as

\text{y}-\text{y}_0=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2

Substituting the given values, we have

\begin{align*} \text{y}-\text{y}_0 &=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2 \\ \text{H}-1.5\:\text{m} & =\left(30\:\text{m/s}\:\sin 60^{\circ} \right)\left(4.0\:\text{s}\right)+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right)^2 \\ \text{H}-1.5\:\text{m} & =25.44\:\text{m} \\ \text{H} & =25.44\:\text{m}+1.5\:\text{m} \\ \text{H} & =26.94\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Part B

The maximum height of the projectile is given by the formula

\Delta \text{y}=\frac{\text{v}_{\text{0y}}^2}{2\text{g}}

or the formula can be written as

\text{y}_{\text{max}}-\text{y}_{\text{0}}=\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}

Therefore, we have

\begin{align*} \text{y}_{\text{max}}-\text{y}_{\text{0}} & =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}} \\ \text{y}_{\text{max}}& =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}+\text{y}_{\text{0}} \\ \text{y}_{\text{max}}&=\frac{\left(\left(30\:\text{m/s}\right)\sin 60^{\circ} \right)^2}{-2\left(-9.81\:\text{m/s}^2\right)}+1.5\:\text{m} \\ \text{y}_{\text{max}}&=34.40\:\text{m}+1.5\:\text{m} \\ \text{y}_{\text{max}} & =35.90\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

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Part C

To solve for the final speed, we need the vertical and horizontal components when the arrow hits the cliff.

We are going to use the formula for acceleration along the vertical to solve for the final speed in the vertical direction. That is

\text{a}=\frac{\text{v}_{\text{y}}-\text{v}_{\text{0y}}}{\text{t}}

Solving for v_fy in terms of the other variables, we have

\begin{align*} \text{v}_{\text{y}}&=\text{v}_{\text{0y}}+\text{at}\\ \text{v}_{\text{y}}&=\left(30\:\text{m/s}\right)\sin 60^{\circ} +\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right) \\ \text{v}_{\text{y}}&=-13.25\:\text{m/s} \end{align*}

Since we know that the horizontal component of the velocity does not change along the entire flight, we can equate the initial and final horizontal velocities. That is

\begin{align*} \text{v}_{\text{x}}&=\text{v}_{\text{0x}}=\left(30\:\text{m/s}\right)\cos 60^{\circ} \\ \text{v}_{\text{x}}&=15\:\text{m/s} \end{align*}

Therefore, the final speed is

\begin{align*} \text{v}&=\sqrt{\text{v}_{\text{y}}^2+\text{v}_{\text{x}}^2} \\ \text{v}&=\sqrt{\left(-13.25\:\text{m/s}\right)^2+\left(15\:\text{m/s}\right)^2}\\ \text{v}&=20.01\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Tag Archives: Physics Solutions

College Physics by Openstax Chapter 2 Problem 52

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

Skidding on a Curve: Unbanked Curves – Uniform Circular Motion Example Problem

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is $\mu _s=0.60$ ; (b) the pavement is icy and $\mu _s=0.25$ .

Acceleration of a Revolving Ball – Uniform Circular Motion Example

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

College Physics 2.51 – Time of the hiker to move out from a falling rock

College Physics 2.50 – Motion of a Jumping Kangaroo

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Physics

College Physics by Openstax Chapter 3 Problem 37

College Physics by Openstax Chapter 3 Problem 36

The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

College Physics by Openstax Chapter 3 Problem 35

College Physics by Openstax Chapter 3 Problem 34

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is \mu _s=0.60; (b) the pavement is icy and \mu _s=0.25.

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is $\mu _s=0.60$ ; (b) the pavement is icy and $\mu _s=0.25$ .