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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 101: Lesson Check

Exercise 55

Step 1

1 of 2

For all the objects, the acceleration due to gravity is the same, irrespective of the weight or shape of an object, assuming no air friction. Hence, on the Moon (an example of vacuum) we expect that the acceleration of the heavier rock will be the same as the acceleration of the lighter rock.

Result

2 of 2

Click here to see the explanation.

Exercise 56

Step 1

1 of 2

In this problem, we explain why a person doing cannon ball dive is an example of free-fall while a person descending on a parachute is not.

Step 2

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For an object to be in free-fall, the only force it experiences is the gravitational force of the Earth. For the cannonball dive, there is no other force aside from gravitational force. For the parachutist, however, the air exerts force on the parachute, hence there are other forces acting on the person. This means that the parachutist is not free-falling.

Exercise 57

Step 1

1 of 2

In this problem, we find the common property of all free-falling objects.

Step 2

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All free falling objects experience only one force in their trajectory – the gravitational force of the Earth on the object. They must be traveling with constant downward acceleration due to gravity.

Exercise 58

Step 1

1 of 2

If we assume no friction then both the gloves (either dropped or tossed) have same acceleration due to gravity and each will be considered to be a freely falling object. Hence, the acceleration of Zoe’s glove is equal to the acceleration of Mickey’s glove; both accelerations are $9.81 m/s^2$ downward.

Result

2 of 2

Click here for the explanation.

Exercise 59

Step 1

1 of 3

In this problem, a girl on a trampoline jump straight upwards with velocity $v_text{i} = 4.5~mathrm{m/s^{2}}$. We find her velocity when she returns to the trampoline.

Step 2

2 of 3

In this case, the displacement is 0, so the magnitude of the velocity is the same. However, the velocity is now pointing downwards, so her velocity when she returns to the trampoline must be negative of her initial velocity,
$$
boxed{ v_text{f} = -4.5~mathrm{m/s} }
$$

Result

3 of 3

$$
v_text{f} = -4.5~mathrm{m/s}
$$

Exercise 60

Step 1

1 of 2

$$
tt{using the Velocity-Position equation $v_{f}^2=v_{i}^2+2aDelta x$ we can determine the velocity of the shell when it hit the rocks, we will choose the coordinate system origin at the point where the motion originates and the positive direction downwards $Rightarrow a=+g,Delta x=+14$:}
$$

$$
begin{align*}
v_{f}^2&=0+2gDelta x\
v_f&=sqrt{2g Delta x}\
&=sqrt{2*(9.81frac{m}{s^2})*(14m)}\
&=boxed{textcolor{#4257b2}{16.6frac{m}{s}}}
end{align*}
$$

Result

2 of 2

$$
tt{$16.6frac{m}{s}$}
$$

Exercise 61

Solution 1

Solution 2

Step 1

1 of 5

In this problem, a volcano launches a lava bomb straight upwards with initial speed $v_text{i} = 28~mathrm{m/s}$. Taking upwards to be the positive direction we find the speed and direction of motion of the lava bomb after (a) $t = 2.0~mathrm{s}$ and (b) $t = 3.0~mathrm{s}$ after launch. We use $g = 9.81~mathrm{m/s^{2}}$.

Step 2

2 of 5

The upward direction is positive, so the acceleration due to gravity is pointing downward (negative). The initial velocity is upward, so $v_text{i} = +28~mathrm{m/s}$. The position time equation of the lava bomb is

$$
begin{align*}
v &= v_text{i} + at \
v &= v_text{i} – gt tag{1}
end{align*}
$$

Step 3

3 of 5

Part A.

For this part, we substitute $t = 2.0~mathrm{s}$ to equation (1).

$$
begin{align*}
v &= v_text{i} – gt \
&= +28~mathrm{m/s} – left( 9.81~mathrm{m/s^{2}} right) left( 2.0~mathrm{s} right) \
&= +8.38~mathrm{m/s} \
v &= boxed{ +8.4~mathrm{m/s} }
end{align*}
$$

The positive sign tells us that the object is moving upwards.

Step 4

4 of 5

Part B.

For this part, we substitute $t = 3.0~mathrm{s}$ to equation (1).

$$
begin{align*}
v &= v_text{i} – gt \
&= +28~mathrm{m/s} – left( 9.81~mathrm{m/s^{2}} right) left( 3.0~mathrm{s} right) \
&= -1.43~mathrm{m/s} \
v &= boxed{ -1.4~mathrm{m/s} }
end{align*}
$$

The negative sign tells us that the object is moving downwards.

Result

5 of 5

begin{enumerate}
item [(a)] $v = +8.4~mathrm{m/s}$
item [(b)] $v = -1.4~mathrm{m/s}$
end{enumerate}

Step 1

1 of 5

To compute the speed and direction of the lava bomb, first, we can use the equation

$$
v_{f} = v_{i} + at
$$

Step 2

2 of 5

According to the problem, upward direction is the positive direction therefore,

$$
a = -g = -9.81m/s^2
$$

Step 3

3 of 5

Substituting values when (a) $t = 2.0s$

$$
v_{f} = 28m/s + (-9.81m/s^2)(2.0s) = 8.4m/s
$$

since the result is positive, the direction of the motion is upward.

Step 4

4 of 5

Substituting values when (b) $t = 3.0s$

$$
v_{f} = 28m/s + (-9.81m/s^2)(3.0s) = -1.43m/s
$$

since the result is negative, the direction of the motion is downward.

Result

5 of 5

a. $speed = 8.4m/s$ , upward direction

b. $speed = 1.43m/s$, downward direction

Exercise 62

Step 1

1 of 4

In this problem, Bernardo steps off at a $x_text{i, B} = 3.0~mathrm{m}$ high diving board. At the same time, Michi jump off with velocity $v_text{i, M} = +4.2~mathrm{m/s}$ from a diving board of height $x_text{i, M} = 1.0~mathrm{m}$. We write the position-time direction of Bernardo and Michi, with upwards being the positive $x$ direction. We use $g = 9.81~mathrm{m/s^{2}}$

Step 2

2 of 4

The position-time equation is

$$
begin{align*}
x &= x_text{i} + v_text{i}t + frac{1}{2}at^{2} tag{1}
end{align*}
$$

In this situation, since acceleration due to gravity is downwards, $a = -9.81~mathrm{m/s^{2}}$.

Step 3

3 of 4

For Bernardo, he just steps out of the board so $v_text{i, B} = 0$. The position-time equation is

$$
begin{align*}
x_{B} &= x_text{i, B} + v_text{i, B}t + frac{1}{2}at^{2} \
&= 3.0~mathrm{m} + 0 – frac{9.81~mathrm{m/s^{2}}}{2}t^{2} \
x_{B} &= boxed{ 3.0~mathrm{m} – left( 4.905~mathrm{m/s^{2}} right)t^{2} }
end{align*}
$$

Step 4

4 of 4

For Michi, we have

$$
begin{align*}
x_{M} &= x_text{i, M} + v_text{i, M}t + frac{1}{2}at^{2} \
&= 1.0~mathrm{m} + left( 4.2~mathrm{m/s} right)t – frac{9.81~mathrm{m/s^{2}}}{2}t^{2} \
x_{M} &= boxed{ 1.0~mathrm{m} + left( 4.2~mathrm{m/s} right)t – left( 4.905~mathrm{m/s^{2}} right)t^{2} }
end{align*}
$$

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