Home Page Textbooks Physics: Principles and Problems Page 255: Standardized Test Practice

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

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Page 255: Standardized Test Practice

Exercise 1

Step 1

1 of 3

A star near the and of its lifetime begins to collapse while still rotate. To find out if the moment of inertia changes and how, we can use this equation:

$$
I=sum_i^nm_ir_i^2
$$

Angular momentum has to be conserved if there is no external torque.

And to find out if angular velocity changes we can use equation for conservation of angular momentum:

$$
L=Iomega
$$

Step 2

2 of 3

Because in this case star collapse it means that each part of the star has smaller $r$ and we can conclude that $textbf{moment of inertia decrease}$.

In this case there is no external forces, that means that angular momentum is conserved and we can conclude that angular momentum $textbf{stays constant}$.

We know that angular momentum is constant and that moment of inertia decrease, hence angular velocity $textbf{increase}$.

Answer is $textbf{B}$.

Result

3 of 3

Exercise 2

Step 1

1 of 4

A ice-skater glides towards a sled and holds on it, they continue sliding in the same direction. It is given that ice-skater has mass of 40 kg and original speed of 2 m/s and sled has mass of 10 kg. We have to find speed of the ice-skater and the sled after collision.

Step 2

2 of 4

To solve this you can use equation for conservation of momentum, because momentum before release and after has to be the same.

$$
begin{align*}
m_1v_1+m_2v_2=m_1v_1’+m_2v_2’\
end{align*}
$$

Step 3

3 of 4

When we put numbers in we get:

$$
begin{align*}
40cdot2+10cdot0&=40v’+10v’\
v’&=frac{40cdot2}{40+10}
end{align*}
$$

$$
boxed{v’=1.6,,rm m/s}
$$

The speed of the ice-skater and the sled after they collide is 1.6 m/s and the answer is $textbf{C}$.

Result

4 of 4

Exercise 3

Step 1

1 of 4

A bicyclist slows down by applying the brakes. We have to calculate the angular impulse in each wheel. It is given that the angular momentum of each wheel decreases form $7,,rm kgm^2/s$ to $3.5,,rm kgm^2/s$ over 5 s.

Step 2

2 of 4

Change in angular momentum is equal to the angular impulse. We just have to calculate change in angular momentum:

$$
Delta L=L_2-L_1
$$

Step 3

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When we put numbers in we get:

$$
Delta L=3.5-7
$$

$$
boxed{Delta L=-3.5,,rm kgm^2/s}
$$

The angular impulse on each wheel is $-3.5,,rm kgm^2/s$, the answer is $textbf{D}$.

Result

4 of 4

Exercise 4

Step 1

1 of 4

A skater stands et the rest on the ice and his friend tosses him a ball, after catching the ball he moves backwards. We have to calculated speed of the ball at the moment just before the skater caught it. It is also given that skater has mass of 45 kg and ball has mass of 5 kg and they have speed of 0.5 m/s after collision.

Step 2

2 of 4

To solve this you can use equation for conservation of momentum, because momentum before release and after has to be the same.

$$
begin{align*}
m_1v_1+m_2v_2=m_1v_1’+m_2v_2’\
end{align*}
$$

Step 3

3 of 4

When we put numbers is we get:

$$
begin{align*}
45cdot0+5v_2&=45cdot0.5+5cdot0.5\
v_2&=frac{45cdot0.5+5cdot0.5}{5}
end{align*}
$$

$$
boxed{v_2=5,,rm m/s}
$$

The speed of the ball at the moment just before the skater caught it is 5 m/s. The answer is $textbf{D}$.

Result

4 of 4

Exercise 5

Step 1

1 of 4

We have to find difference in momentum between runner and a truck. It is given that mass of a runner is 50 kg and his speed is 3 m/s and mass of a truck is $3cdot10^{3},,rm kg$ and his speed is 1 m/s.

Step 2

2 of 4

This can be calculated with this simple equation:

$$
p_{t}-p_{r}=m_tv_t-m_rv_r
$$

Step 3

3 of 4

When we put numbers we get:

$$
p_t-p_r=3cdot10^{3}cdot1-50cdot3
$$

$$
boxed{p_t-p_r=2850,,rm kgm/s}
$$

The difference in momentum is $2850,,rm kgm/s$, the answer is $textbf{C}$.

Result

4 of 4

Exercise 6

Step 1

1 of 4

Two gears are in contact, the larger gear has twice the radius and four times the mass of the smaller gear. We have to calculate what is the angular momentum of the larger gear as a function of the angular momentum of the smaller gear. We know that moment of inertia of a disk is $frac{1}{2}mr^2$.

Step 2

2 of 4

An angular momentum of a gear is:

$$
L=Iomega
$$

we know moment of inertia, so we just put that in, but in this case we have to put angular velocity in terms of linear speed because linear velocity is the same for both gears but in opposite direction.

$$
L=frac{1}{2}mr^2cdotfrac{v}{r}
$$

We can calculate ration with equaiton:

$$
begin{align*}
frac{L_l}{L_s}&=frac{frac{1}{2}m_lr_lv_l}{frac{1}{2}m_sr_sv_s}\
frac{L_l}{L_s}&=frac{m_lr_lv_l}{m_sr_sv_s}\
end{align*}
$$

Step 3

3 of 4

When we put number is we get:

$$
frac{L_l}{L_s}=frac{4m_scdot2r_scdot1v_s}{m_sr_sv_s}
$$

$$
L_l=-8L_s
$$

The answer is $textbf{C}$.

Result

4 of 4

Exercise 7

Step 1

1 of 4

The rock fly off the ground because a force is exerted on it. We have to calculate the mass of the rock. It is given that force is 16 N, an impulse is 0.8 kgm/s and speed of a rock after is 4 m/s.

Step 2

2 of 4

We know that change in momentum is equal to impulse. Before the force is exerted momentum of the rock is zero so we can write:

$$
Ft=mvrightarrow m=frac{Ft}{v}
$$

Step 3

3 of 4

When we put numbers in we get:

$$
m=frac{0.8}{4}
$$

$$
boxed{m=0.2,,rm kg}
$$

The answer is $textbf{A}$.

Result

4 of 4

Exercise 8

Step 1

1 of 4

A rock falls to the ground. We have to calculate the impulse on the rock. It is given that velocity at the moment it strikes the ground is 20 m/s and its mass is 12 kg.

Step 2

2 of 4

The impulse is equal to change in momentum. After it hits the ground its momentum is zero and we can write the equation:

$$
Ft=p_2-p_1=-mv
$$

Step 3

3 of 4

When we put number is we get:

$$
Ft=-12cdot20
$$

$$
boxed{Ft=-240,,rm kgm/s}
$$

Result

4 of 4

$$
Ft=-240,,rm kgm/s
$$

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Page 255: Standardized Test Practice

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