Home Page Textbooks Physics: Principles and Problems Page 169: Standardized Test Practice

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

Textbook solutions

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Page 169: Standardized Test Practice

Exercise 1

Step 1

1 of 2

We should find the flight time:

$y = (-1/2) g t^2 + v_i sin(theta) t$

$-1.60 = (-1/2) (9.80) t^2 + (9.40) sin(41.0) t$

$==> t = 1.479 s$

The horizontal distance is:

$$
x = v_x t = (v_i cos(41.0)) t = (9.40 cos(41.0)) (1.479) = 10.5 m
$$

Result

2 of 2

10.5 m

Exercise 2

Solution 1

Solution 2

Step 1

1 of 2

$a_c = dfrac{v^2}{r}$

$a_c = dfrac{0.89^2}{2.8}$

$$
a_c = 0.28 m/s^2
$$

Result

2 of 2

$$
0.28 m/s^2
$$

Step 1

1 of 3

**Given:**
– Radius: $r = 2.8 mathrm{~m}$;
– Velocity: $v = 0.89 ,frac{text{m}}{text{s}}$;

**Required:**
– The acceleration $a_text c$;

Step 2

2 of 3

Centripetal acceleration is defined as the acceleration in uniform circular motion. It can be calculated as the ratio of tangential velocity squared and radius.
$$begin{align*}
a_text c &= frac{v^2}{r} \
&= dfrac{ left( 0.89 ,frac{text{m}}{text{s}} right)^2}{2.8 mathrm{~m}} \
&= 0.28 ,frac{text{m}}{text{s}^2}
end{align*}$$
$$boxed{ a_text c = 0.28 ,frac{text{m}}{text{s}^2} }$$

Result

3 of 3

$$a_text c = 0.28 ,frac{text{m}}{text{s}^2}$$

Exercise 3

Step 1

1 of 2

The centripetal acceleration is:

$a_c = dfrac{F}{m} = dfrac{4.0}{0.82} = 4.878 m/s^2$

The equation of the centripetal acceleration is:

$a_c = dfrac{v^2}{r}$

Thus:

$$
v = sqrt{r a_c} = sqrt{(2.0)*(4.878)} = 3.1 m/s
$$

Result

2 of 2

$$
3.1 m/s
$$

Exercise 4

Step 1

1 of 2

The equation of the centripetal acceleration is:

$a_c = dfrac{v^2}{r} = dfrac{(20.0)^2}{80.0} = 5.00 m/s^2$

The force is:

$F = m a = (1000) (5.00) = 5.0 times 10^3 N$

Result

2 of 2

$$
5.0 times 10^3 N
$$

Exercise 5

Step 1

1 of 2

$v = 10 + 20 = 30 km/h$

We should convert km/h to m/s:

$$
v = (30 km/h) (dfrac{1 m/s}{3.6 km/h}) = 8 m/s
$$

Result

2 of 2

$$
8 m/s
$$

Exercise 6

Solution 1

Solution 2

Step 1

1 of 2

$y = dfrac{v_f^2 – vi^2}{- 2 g} = dfrac{(0.0)^2 – (18 sin(78))^2}{-2 times 9.8} = 16 m$

Result

2 of 2

$$
16 m
$$

Step 1

1 of 7

**Given:**
– Mass: $m = 125 mathrm{~g}$;
– Angle: $alpha = 78°$;
– Velocity: $v_0 = 18 ,frac{text{m}}{text{s}}$;

**Required:**
– The maximum height an apple reaches $y$;

Step 2

2 of 7

The apple is thrown out under the angle with initial speed. We call that projectile motion. In projectile motion, the horizontal component of speed remains constant, and the vertical component of speed is changing due to gravity.
$$begin{align*}
{v_text y}^2 &= {v_{0, text y}}^2 + 2gy
end{align*}$$

Step 3

3 of 7

The problem looks like this:

![‘slader’](https://slader-solution-uploads.s3.amazonaws.com/d1eeb24c-a809-43fb-afcc-6453a0461274-1649508306917378.png)

Step 4

4 of 7

The apple is thrown with initial velocity under the angle. That means we can rewrite it in terms of components. To do so, we use basic trigonometry identities. From the picture above it is seen that the component in the $y$ direction is opposite to the angle $alpha$, hence we will use the definition of sine and calculate the magnitude of the component as:
$$begin{align*}
sin alpha &= frac{ v_{0, text y} }{ v_0} \
v_{0, text y} &= v_0 cdot sin alpha \
&= 18 ,frac{text{m}}{text{s}} cdot sin 78° \
&= 17.61 ,frac{text{m}}{text{s}}
end{align*}$$

Step 5

5 of 7

Now we can use the given equation to find the maximum height an apple reaches. Solving for $y$ we subtract ${v_{0, text y}}^2$ and divide it by $2g$:
$$ y = frac{{v_text y}^2 – {v_{0, text y}}^2 }{2g} $$

Step 6

6 of 7

Keep in mind that at the highest point the apple stops and its speed $v_text y$ is zero, and that acceleration is in the negative $y$ direction. Putting everything together, we have:
$$begin{align*}
y &= frac{- {v_{0, text y}}^2 }{2g} \

&= frac{- left( 17.61 ,frac{text{m}}{text{s}} right)^2}{2 cdot left( – 9.8 ,frac{text{m}}{text{s}^2} right) } \

&= 15.8 mathrm{~m} \
&approx 16 mathrm{~m}
end{align*}$$
$$boxed{ y approx 16 mathrm{~m} }$$

Result

7 of 7

$$y approx 16 mathrm{~m} $$

Exercise 7

Solution 1

Solution 2

Step 1

1 of 2

D) The two objects will hit the ground at the same time.

Result

2 of 2

Step 1

1 of 5

**a)** This answer is not correct because the acceleration due to gravity doesn’t depend on the object’s mass.

Step 2

2 of 5

**b)** The speed the object is moving at doesn’t affect the acceleration due to gravity.

Step 3

3 of 5

**c)** This isn’t true because the initial speed of the objects is not the same. Even though they are both accelerated by the same acceleration, their final velocities won’t be the same.

Step 4

4 of 5

**d) Because the same acceleration acts upon both objects, they will still hit the ground at the same time, under the condition that they are dropped from the same height.**

Result

5 of 5

$$text{d)}$$

Exercise 8

Step 1

1 of 2

The flight time is:

$t = sqrt{dfrac{-2y}{g}} = sqrt{dfrac{(-2)(52)}{9.8}} = 3.258 s$

The horizontal displacement of the ball is:

$x = v_x t = (25) (3.258) = 81 m$

Thus the ball will past the ring.

Result

2 of 2

Yes, they will.

Exercise 9

Step 1

1 of 2

The equation of the centripetal acceleration is:

$a_c = dfrac{4 pi^2 r}{T^2} = dfrac{(4)*(3.14)^2*(0.86)}{(1.8)^2} = 10.5 m/s^2$

The tension in the magical chain is:

$$
F = m a = (5.6) (10.5) = 59 N
$$

Result

2 of 2

$$
59 N
$$

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Page 169: Standardized Test Practice

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