(force),(time)$rightarrow$(impulse)$rightarrow$(momentum)$leftarrow$(mass),(velocity)

Momentum depends on mass and velocity:

$$
p=mv
$$

The answer is $textbf{yes}$, because it has smaller mass its velocity has to be bigger.

If acceleration is 0, then velocity will be constant. Hence the momentum will be constant

$$
text{color{#c34632}Remember that : Linear momentum = Mass $times$ Velocity}
$$

Notice that this is true only about linear momentum, angular momentum can change even if the net force on the system is 0

$$
text{color{#4257b2}No change in Linear momentum can occur if net force is 0}
$$

Yes, object can obtain a larger impulse from smaller force if that force is acting on it for longer period of time.

$$
impulse=Ft
$$

You should move your hands in the same direction as the moving ball, because to stop the ball you have to change its momentum and you want to do it in longer period of time so that average force is smaller.

$$
Delta p=Ft
$$

A)

After impulse A an object has an change in momentum:

$$
Delta p=Ft=2cdot2=4,,rm kgm/s
$$

during that time an object accelerate.

B)

After impulse A an object has an change in momentum:

$$
Delta p=Ft=-2cdot2=-4,,rm kgm/s
$$

during that time an object deaccelerate.

C)

After impulse A an object has an change in momentum:

$$
Delta p=Ft=-2cdot=-2,,rm kgm/s
$$

during that time an object accelerate but in opposite direction.

If we use conservation of momentum and assume that two trucks are identical we would get:

$$
mv=2mv’rightarrow v’=frac{v}{2}
$$

But they are moving more than half the speed of the moving truck and that means that moving truck must have bigger mass.

Impulse is equal to change in momentum:

$$
begin{align*}
Ft&=Delta p\
Ft&=mv\
t&=frac{mv}{F}\
&=frac{0.058cdot62}{272}
end{align*}
$$

$$
boxed{t=0.013,,rm s}
$$

$$
t=0.013,,rm s
$$

a)

Change in momentum is equal to:

$$
begin{align*}
Delta p&=mv’-mv\
&=0.145cdot(-58)-0.145cdot42
end{align*}
$$

$$
boxed{Delta p=-14.5,,rm kgm/s}
$$

b)

Impulse is equal to the change in momentum:

$$
begin{align*}
Ft&=Delta p\
F&=frac{Delta p}{t}\
&=frac{-14.5}{4.6cdot10^{-4}}
end{align*}
$$

$$
boxed{F=-31522,,rm N}
$$

a) $Delta p=-14.5,,rm kgm/s$

b) $F=-31522,,rm N$

Given values from the problem are the force $F$ acting upon the bowling ball, the mass $m$ of the bowling ball, and the time $Delta{t}$ duration in which there is force being acted upon the bowling ball.

$$
begin{align*}
F &= 186 ,mathrm{N}\
m &= 7.3 ,mathrm{kg} \
Delta{t} &= 0.40 ,mathrm{s}
end{align*}
$$

$textbf{Change in Momentum.}$

Given the formula such that $Delta{p}$ is the change in momentum, $F$ is the force acting upon the object, and time $t$ is the duration in which the force is acting upon the same object.

$$
begin{equation}
Delta{p} = FDelta{t}
end{equation}
$$

We can solve for the momentum by plugging in the given values.

$$
begin{align*}
Delta{p} &= FDelta{t}\
&= (186,mathrm{N})(0.40,mathrm{s})\
&= 74 N cdot s\
&= boxed{74 ,mathrm{kg}cdot mathrm{frac{m}{s}}}
end{align*}
$$

$$
textbf{Change in Velocity.}
$$

Given the formula where $Delta{v}$ is the change in velocity, $Delta{p}$ is the change in momentum, and $m$ is the mass.

$$
begin{equation}
Delta{v} = frac{Delta{p}}{m}
end{equation}
$$

We can solve for the momentum by using $Delta{p}$ acquired from the previous solution and the mass($m$) in the given values.

$$
begin{align*}
Delta{v} &= frac{Delta{p}}{m}\
&= frac{74 ,mathrm{kg}cdot mathrm{frac{m}{s}}}{7.3,mathrm{kg}}\
&= boxed{10 ,mathrm{frac{m}{s}}}
end{align*}
$$

$Delta{p} = 74 ,mathrm{kg}cdot mathrm{frac{m}{s}}$ and $Delta{v} = 10 ,mathrm{frac{m}{s}}$

Given: $m=5500,,rm kg$, $v_1=4.2,,rm m/s$, $v_2=7.8,,rm m/s$, $t=15,,rm s$

a)

We can calculate change in momentum:

$$
Delta p=mv_2-mv_1=5500cdot7.8-5500cdot4.2
$$

$$
boxed{Delta p=19800,,rm kgm/s}
$$

b)

Impulse is equal to change in momentum:

$$
begin{align*}
Ft&=Delta p\
F&=frac{Delta p}{t}\
&=frac{19800}{15}
end{align*}
$$

$$
boxed{F=1320,,rm N}
$$

a) $Delta p=19800,,rm kgm/s$

b) $F=1320,,rm N$

Given: $m=6,,rm g$, $v=350,,rm m/s$, $t=1.8,,rm ms$

Impulse is equal to change in momentum:

$$
begin{align*}
Ft&=mv\
F&=frac{mv}{t}\
&=frac{6cdot10^{-3}cdot350}{1.8cdot10^{-3}}
end{align*}
$$

$$
boxed{F=1167,,rm N}
$$

$$
F=1167,,rm N
$$

Given: $m=0.24,,rm kg$, $v_1=3.8,,rm m/s$, $v_2=-2.4,,rm m/s$, $t=0.025,,rm s$

Impulse is equal to change in momentum:

$$
begin{align*}
Ft&=mv_2-mv_1\
F&=frac{mv_2-mv_1}{t}\
&=frac{0.24cdot(-2.4)-0.24cdot3.8}{0.025}
end{align*}
$$

$$
boxed{F=59.5,,rm N}
$$

$$
F=59.5,,rm N
$$

The impulse given to the puck is:

$$
I = Delta p = F Delta t = (30.0) (0.16) = 4.8 Kg.m/s
$$

$$
4.8 Kg.m/s
$$

Given: $m_1=35.6,,rm kg$, $m=1.3,,rm kg$, $v=9.5,,rm m/s$

Combined momentum is equal to:

$$
begin{align*}
p&=(m_1+m_2)v\
&=(35.6+1.3)cdot9.5
end{align*}
$$

$$
boxed{p=350.6,,rm kgm/s}
$$

$$
p=350.6,,rm kgm/s
$$

The impulse given to the puck increases its momentum:

$p = F t = (30.0) (0.16) = 4.8 Kg.m/s$

Its speed is:

$$
v = dfrac{p}{m} = dfrac{4.8}{0.115} = 42 m/s
$$

$$
42 m/s
$$

Given: $m=25,,rm kg$, $v_1=12,,rm m/s$, $v_a=8,,rm m/s$, $v_b=-8,,rm m/s$

a)

Impulse is equal to change in momentum:

$$
begin{align*}
Ft&=mv_a-mv_1\
&=25cdot8-25cdot12\
end{align*}
$$

$$
boxed{Ft=-100,,rm kgm/s}
$$

b)

Impulse is equal to change in momentum:

$$
begin{align*}
Ft&=mv_a-mv_1\
&=25cdot(-8)-25cdot12\
end{align*}
$$

$$
boxed{Ft=-500,,rm kgm/s}
$$

a) $Ft=-100,,rm kgm/s$

b) $Ft=-500,,rm kgm/s$

Given: $m=0.15,,rm kg$, $v_1=12,,rm m/s$

Area under the graph is change in momentum:

$$
begin{align*}
Delta p&=2cdotleft(frac{2cdot1}{2}right)\
Delta p&=2,,rm kgm/s
end{align*}
$$

now we can calculate speed:

$$
mv_1+Delta p=mvrightarrow v=frac{mv_1+Delta p}{m}=frac{0.15cdot12+2}{0.15}
$$

$$
boxed{v=25,,rm m/s}
$$

$$
v=25,,rm m/s
$$

Given: $m=0.145,,rm kg$, $v=35,,rm m/s$

a)

Change in momentum is equal to:

$$
Delta p=mv=0.145cdot35
$$

$$
boxed{Delta p=5.075,,rm kgm/s}
$$

b)

Impulse is equal o change in momentum:

$$
begin{align*}
Ft&=Delta p\
F&=frac{Delta p}{t}\
&=frac{5.075}{0.05}
end{align*}
$$

$$
boxed{F=101.5,,rm N}
$$

b)

Impulse is equal o change in momentum:

$$
begin{align*}
Ft&=Delta p\
F&=frac{Delta p}{t}\
&=frac{5.075}{5}
end{align*}
$$

$$
boxed{F=10.15,,rm N}
$$

a) $Delta p=5.075,,rm kgm/s$

b) $F=101.5,,rm N$

c) $F=10.15,,rm N$

Given: $m=0.115,,rm kg$, $v_1=37,,rm m/s$, $v_2=-25,,rm m/s$

a)

Impulse is equal to change in momentum:

$$
begin{align*}
Ft&=mv_2-mv_1\
&=0.115cdot(-25)-0.115cdot37
end{align*}
$$

$$
boxed{Ft=-7.13,,rm kgm/s}
$$

b)

Average force is equal to:

$$
F=frac{Ft}{t}=frac{-7.13}{5cdot10^{-4}}
$$

$$
boxed{F=-14260,,rm N}
$$

a) $Ft=-7.13,,rm kgm/s$

b) $F=-14260,,rm N$

a)

To calculate impulse we can use difference in momentum:

$$
begin{align*}
Ft&=p_2-p_1\
&=mv_2-mv_1\
&=4.7cdot10^{-26}cdot(-550)-4.7cdot10^{-26}cdot550
end{align*}
$$

$$
boxed{Ft=-5.17cdot10^{-23},,rm kgm/s}
$$

We put minus sign on $v_2$ because it is in opposite direction.

b)

We can calculate average force by using the definition of impulse, for time duration we will put one second and for number of collisions we will put the number of collisions in one second:

$$
begin{align*}
F&=nfrac{Ft}{t}\
&=1.5cdot10^{23}frac{-5.17cdot10^{-23}}{1}
end{align*}
$$

$$
boxed{F=-7.76,,rm N}
$$

a) $Ft=-5.17cdot10^{-23},,rm kgm/s$

b) $F=F=-7.76,,rm N$

To calculate time that should it be fired we can use that impulse is equal to change in momentum:

$$
begin{align*}
Ft&=mv\
t&=frac{mv}{F}\
&=frac{72000cdot63cdot10^{-2}}{35}
end{align*}
$$

$$
boxed{t=1296,,rm s}
$$

$$
t=1296,,rm s
$$

To calculate vertical velocity of a bale of hay we can use conservation of energy:

$$
begin{align*}
mgh&=frac{mv_v^2}{2}\
v_v&=sqrt{2gh}\
v_v&=sqrt{2cdot9.81cdot60}\
v_v&=34.3,,rm m/s
end{align*}
$$

We can get mass of hey from weight:

$$
begin{align*}
F_g&=mg\
m&=frac{F_g}{g}\
m&=frac{175}{9.81}\
m&=17.85,,rm kg
end{align*}
$$

Now we can calculate magnitude of the momentum in vertical direction:

$$
begin{align*}
p_v&=mv_v\
p_v&=17.85cdot1177.2\
p_v&=612.4,,rm kgm/s
end{align*}
$$

Horizontal velocity is the same as planes:

$$
v_h=36,,rm m/s
$$

And magnitude of momentum in horizontal direction is:

$$
begin{align*}
p_h&=mv_v\
p_h&=17.85cdot36\
p_h&=642.2,,rm kgm/s
end{align*}
$$

Now we can calculate magnitude of total momentum:

$$
begin{align*}
p&=sqrt{p_v^2+p_h^2}\
p&=sqrt{612.4^2+642.2^2}
end{align*}
$$

$$
boxed{p=887.4,,rm kgm/s}
$$

We can calculate direction with equation:

$$
begin{align*}
tanphi=frac{p_v}{p_h}\
tanphi=frac{612.4}{642.2}
end{align*}
$$

$$
boxed{phi=44rm^o,,, towards,, the,, center,, of,, the,, earth}
$$

$$
p=887.4,,rm kgm/s
$$

$$
phi=44rm^o
$$

a)

Impulse is equal to change in momentum:

$$
begin{align*}
Ft&=mv\
Ft&=20cdot10
end{align*}
$$

$$
boxed{Ft=200,,rm kgm/s}
$$

b)

Now when we have impulse we can get average force because impulse is equal to force times time:

$$
begin{align*}
F=frac{Ft}{t}\
F=frac{200}{0.05}
end{align*}
$$

$$
boxed{F=4000,,rm N}
$$

c)

We can calculate mass of an object whose weight equals the force with equation:

$$
begin{align*}
F_g&=mg\
m&=frac{F_g}{g}\
m&=frac{4000}{9.81}
end{align*}
$$

$$
boxed{m=408,,rm kg}
$$

d)

World record in lifting weight is 501 kg in deadlift, in that movement human uses almost every muscle, we can conclude that using just arm human $textbf{can not lift that weight}$.

e)

Because restraining seat is build to $textbf{prolongs the time}$ in which force acts on the child, hence the average force is smaller.

a) $Ft=200,,rm kgm/s$

b) $F=4000,,rm N$

c) $m=408,,rm kg$

d) No

e) prolongs the time

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

In this case velocity before release is zero for both carts, after the release they have velocity in opposite directions so we have to put minus sign in front of one of them.

$$
begin{align*}
5cdot0+2cdot0&=5cdot0.12-2v_2’\
v_2’&=frac{5cdot0.12}{2}
end{align*}
$$

$$
boxed{v_2’=0.3,,rm m/s}
$$

$$
v_2’=0.3,,rm m/s
$$

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

In this case velocity before launch is the same for both projectile and launcher.

$$
begin{align*}
4.65cdot2+50cdot10^{-3}cdot2&=4.65v_1’+50cdot10^{-3}cdot647\
v_1’&=frac{4.65cdot2+50cdot10^{-3}cdot2-50cdot10^{-3}cdot647}{4.65}
end{align*}
$$

$$
boxed{v_1’=-4.9,,rm m/s}
$$

$$
v_1’=-4.9,,rm m/s
$$

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

$$
begin{align*}
8.5cdot 0+12cdot10^{-3}cdot150&=8.5v_1′-12cdot10^{-3}cdot10^{2}\
v_2’&=frac{12cdot10^{-3}cdot150+12cdot10^{-3}cdot10^{2}}{8.5}
end{align*}
$$

$$
boxed{v_1’=0.35,,rm m/s}
$$

$$
v_1’=0.35,,rm m/s
$$

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

In this case velocity before is the same for Kofi and skateboard, after Kofi jumps skateboard has velocity zero.

$$
begin{align*}
42cdot1.2+2cdot1.2&=42v_1′-2cdot0\
v_1’&=frac{42cdot1.2+2cdot1.2}{42}
end{align*}
$$

$$
boxed{v_1’=1.26,,rm m/s}
$$

The sign of velocity that we got is plus, that means that he jumped in the same direction as he was moving before.

$$
v_1’=1.26,,rm m/s
$$

1.

$$
begin{align*}
m_1v_{x1}+m_2v_{x2}=m_1v_{x1}’+m_2v_{x2}’\
end{align*}
$$

2.

$$
begin{align*}
m_1v_{y1}+m_2v_{y2}=m_1v_{y1}’+m_2v_{y2}’\
end{align*}
$$

Because the angle between $x$ and $y$ component is $45,rm ^o$, they have the same magnitude.

1.

$$
begin{align*}
m_1v_{x1}+m_2cdot0&=m_1v_{x1}’+m_2v_{x2}’\
end{align*}
$$

2.

$$
begin{align*}
m_1v_{y1}’&=m_2v_{y2}’\
m_1v_{x1}’&=m_2v_{x2}’\
end{align*}
$$

Now when we put equation 2. in equation 1. we get:

$$
begin{align*}
m_1v_{x1}&=m_2v_{x2}’+m_2v_{x2}’\
v_{x2}’&=frac{m_1v_{x1}}{2m_2}\
v_{x2}’&=frac{0.16cdot4}{2cdot0.16}\
v_{x2}’&=2,,rm m/s
end{align*}
$$

Now we calculate for other ball:

$$
begin{align*}
m_1v_{x1}&=m_1v_{x1}’+m_1v_{x1}’\
v_{x1}’&=frac{m_1v_{x1}}{2m_1}\
v_{x1}’&=frac{0.16cdot4}{2cdot0.16}\
v_{x1}’&=2,,rm m/s
end{align*}
$$

We know that the angle is 45$rm ^o$ so we can use sinus of that angle to get total velocity.

$$
begin{align*}
sin{45mathrm{^o}}&=frac{v_{x1}’}{v_1′}\
v_1’&=frac{v_{x1}’}{sin{45mathrm{^o}}}
end{align*}
$$

$$
boxed{v_1’=2.8,,rm m/s}
$$

$$
begin{align*}
sin{45mathrm{^o}}&=frac{v_{x2}’}{v_2′}\
v_2’&=frac{v_{x2}’}{sin{45mathrm{^o}}}
end{align*}
$$

$$
boxed{v_2’=2.8,,rm m/s}
$$

$$
v_1’=2.8,,rm m/s
$$

$$
v_2’=2.8,,rm m/s
$$

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

In this case velocity before collision velocity of 825 kg compact car is zero, after the release both cars have the same velocity.

$$
begin{align*}
2575v_1+825cdot0&=2575cdot8.5+825cdot8.5\
v_1&=frac{2575cdot8.5+825cdot8.5}{2575}
end{align*}
$$

$$
boxed{v_1=11.2,,rm m/s}
$$

$$
v_1=11.2,,rm m/s
$$

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

$$
begin{align*}
0.2cdot0.3+0.1cdot0.1&=0.2v_1’+0.1cdot0.26\
v_2’&=frac{0.2cdot0.3+0.1cdot0.1-0.1cdot0.26}{0.2}
end{align*}
$$

$$
boxed{v_1’=0.22,,rm m/s}
$$

$$
v_1’=0.22,,rm m/s
$$

Change in momentum is equal to impulse generated by constant force.

$$
begin{align*}
Ft&=Delta p\
end{align*}
$$

Now we put numbers in and we get that change in momentum is:

$$
begin{align*}
Delta p=6cdot10\
end{align*}
$$

$$
boxed{Delta p=60,,rm kgm/s}
$$

Momentum is defined as mass times velocity, so we can get change in velocity from that.

$$
begin{align*}
Delta p&=p_2-p_1\
Delta p&=mv_2-mv_1\
Delta p&=m(v_2-v_1)\
Delta p&=mDelta v\
Delta v&=frac{Delta p}{m}
end{align*}
$$

Now we put numbers and change in velocity is:

$$
begin{align*}
Delta v=frac{60}{3}
end{align*}
$$

$$
boxed{Delta v=20,,rm m/s}
$$

$$
Delta p=60,,rm kgm/s
$$

$$
Delta v=20,,rm m/s
$$

In part $textbf{a}$ we have to calculate change in momentum of the car and in part $textbf{b}$ we have to calculate magnitude of the force. It is given that mass of the car is 625 kg and that velocity is changed from 10 m/s to 44 m/s in 68 s by external, constant force.

a)

Momentum is defined as mass times velocity, so we can get change in momentum if we know the change in velocity.

$$
begin{align*}
Delta p&=p_2-p_1\
Delta p&=mv_2-mv_1\
Delta p&=m(v_2-v_1)\
end{align*}
$$

Now we can put numbers an we can get change in momentum:

$$
Delta p=625cdot(44-10)
$$

$$
boxed{Delta p=21250,,rm kgm/s}
$$

b)

Change in momentum is equal to impulse generated by constant force.

$$
begin{align*}
Ft&=Delta p\
F&=frac{Delta p}{t}
end{align*}
$$

Now we can put numbers and get average force on the car.

$$
begin{align*}
F=frac{21250}{68}\
end{align*}
$$

$$
boxed{F=312.5,,rm N}
$$

a) $Delta p=21250,,rm kgm/s$

b) $F=312.5,,rm N$

In par $textbf{a}$ we have to calculate change in momentum of the dragster, in part $textbf{b}$ we have to calculate average force on the dragster and in part $textbf{c}$ we have to name what exerts that force.

a)

Momentum is defined as mass times velocity.

$$
begin{align*}
Delta p&=p_2-p_1\
Delta p&=mv_2-mv_1\
Delta p&=m(v_2-v_1)\
end{align*}
$$

Now we can put numbers and we can get change in momenum:

$$
Delta p=845cdot(27.78-0)
$$

$$
boxed{Delta p=23474,,rm kgm/s}
$$

b)

Change in momentum is equal to impulse generated by force.

$$
begin{align*}
Ft&=Delta p\
F&=frac{Delta p}{t}
end{align*}
$$

Now we can put numbers and we can get average force:

$$
F=frac{23474}{0.9}
$$

$$
boxed{F=26082,,rm N}
$$

a) $Delta p=23474,,rm kgm/s$

b) $F=26082,,rm N$

c) race track

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

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

When we put numbers we get that their velocity is:

$$
begin{align*}
0.115cdot35+0.365cdot0&=0.115v’+0.365v’\
v’&=frac{0.115cdot35}{0.115+0.365}
end{align*}
$$

$$
boxed{v’=8.39,,rm m/s}
$$

$$
v’=8.39,,rm m/s
$$

In this problem gymnast is performing a routine in which she does giant swings on the bar holding her body straight, then she lets go of the bar and grabs her knees with her hands, in the final act she straightens up and lands. In part $textbf{a}$ we have to answer around which axis she spin, in part $textbf{b}$ we have to rank, from greatest to least, her momentum of inertia for three positions and in part $textbf{c}$ we have to rank, from greatest to least, here angular velocity for three positions.

b)

Moment of inertia can be calculated with equation:

$$
I=sum_i^nm_ir_i^2
$$

we can see from this equation that moment of inertia is larger the farther mass is from the center of rotation.

Rank in order, from greatest to least, her moments of inertia for the three positions:

$$
boxed{rm first>third>second}
$$

c)

To solve this we have to use conservation of angular momentum, and we know that angular momentum is given by equation:

$$
L=Iomega
$$

We can see form equation that if the angular momentum is conserved, larger the moment of inertia smaller the angular velocity.

Rank in order, from greatest to least, her velocity for the three positions:

$$
boxed{rm second>third>first}
$$

a) center of mass

b) $rm first>third>second$

c) $rm second>third>first$

A male dancer leaps in the air. In part $textbf{a}$ we have to calculate with what momentum does he reach the ground, in part $textbf{b}$ what impulse is needed to stop the dancer, in part $textbf{c}$ we have to find the average force in the dancers body knowing that the stopping time is 0.05 s and in part $textbf{d}$ compare the stopping time with his weight. It is also given that his mass is 60 kg and he leaps 0.32 m high.

a)

To calculate momentum we have to get velocity in that moment, we can do that with equation for conservation of energy:

$$
mgh=frac{mv^2}{2}rightarrow v=sqrt{2gh}
$$

Now when we can get momentum with equation:

$$
p=mv
$$

When we put the numbers in equation for velocity we get:

$$
v=sqrt{2cdot9.81cdot0.32}=2.5,,rm m/s
$$

Now we can get calculate momentum:

$$
p=60cdot2.5
$$

$$
boxed{p=150,,rm kgm/s}
$$

b)

Impulse is equal to change in momentum, in this case we have that starting momentum is that momentum form part $textbf{a}$ and at the end is zero:

$$
Ft=Delta p
$$

$$
boxed{Ft=150,,rm kgm/s}
$$

c)

Impulse is equal to force time time, we know from part $textbf{b}$ impulse and the time is given:

$$
F=frac{Ft}{t}
$$

When we put the numbers we get that average force is:

$$
F=frac{150}{0.05}
$$

$$
boxed{F=3006,,rm N}
$$

d)

Weight of the dancer can be calculated with equation:

$$
F_g=mg
$$

We can compare the stopping force with his weight by dividing them:

$$
x=frac{F}{F_g}
$$

When we put number we get that weight is:

$$
F_g=60cdot9.81=589,,rm N
$$

and finally we get:

$$
x=frac{3006}{589}
$$

$$
boxed{x=5.1}
$$

The stopping force on the body is 5.1 times bigger then weight.

a) $p=150,,rm kgm/s$

b) $Ft=150,,rm kgm/s$

c) $F=3006,,rm N$

d) $x=5.1$

To answer this question we can use conservation of angular momentum. In the starting position when student does not rotate total angular momentum is zero. When student starts to rotate the wheel, wheel gets some angular momentum and if total angular momentum has to be conserved $textbf{student and stool must rotate in the opposite direction}$ then the wheel.

We can calculate centripetal force with equation:

$$
F_{cp}=frac{mv^2}{r}
$$

And centripetal force in this case on the top of the ball’s motion is equal to tension plus weight:

$$
T+mg=frac{mv^2}{r}rightarrow T=frac{mv^2}{r}-mg
$$

When we put the numbers in we get that tension is:

$$
T=frac{0.72cdot3.3^2}{0.6}-0.72cdot9.81
$$

$$
boxed{T=6,,rm N}
$$

$$
T=6,,rm N
$$

We have to calculate the radius of the circular orbit satellite must have to have period of one day. We got hint that we could use Kepler’s third law and that the Moon is $3.9cdot10^8,,rm m$ from earth and its period is 27.33 days.

the third Kepler’s law says that the square of a planet’s orbital period is proportional to the cube of the length of the semi-major axis of its orbit. Because the path of the Moon and the satellite is practically circular, the semi-major axis is equal to radius. Equation for that looks like:

$$
frac{T^2}{T_M^2}=frac{r^3}{r_M^3}rightarrow r=r_Mleft(frac{T^2}{T_M^2}right)^{frac{1}{3}}
$$

When we put numbers we get:

$$
r=3.9cdot10^8cdotleft(frac{1^2}{27.33^2}right)^{frac{1}{3}}
$$

$$
boxed{r=4.3cdot10^{7},,rm m}
$$

$$
r=4.3cdot10^{7},,rm m
$$

A machine pulls a rope that is wrapped around a drum. We have to calculated $alpha$, $omega$ and $I$. It is also given that diameter of the drum is 0.6 m, this means that radius is 0.3 m. Constant force is applied by the machine of 40 N in total of 2 s. And in that time rope unwound 5 m.

To calculate $alpha$ we have to get linear acceleration and we can do that with equation:

$$
x=frac{1}{2}at^2rightarrow a=frac{2x}{t^2}
$$

Now when we have linear acceleration we can get angular acceleration:

$$
alpha=frac{a}{r}
$$

To calculate angular velocity we have to get linear velocity first, we can di that with equation:

$$
v=at
$$

Now it is easy to get angular velocity:

$$
omega=frac{v}{r}
$$

And finally to calculate moment of inertia we have to find torque first:

$$
tau=Fr
$$

now we can get moment of inertia:

$$
tau=Ialpharightarrow I=frac{tau}{alpha}
$$

Lets calculate linear acceleration:

$$
a=frac{2cdot5}{2^2}=2.5,,rm m/s^2
$$

Angular acceleration is:

$$
alpha=frac{2.5}{0.3}
$$

$$
boxed{alpha=8.33,,rm rad/s^2}
$$

Now lets calculate linear velocity:

$$
v=2.5cdot2=5,,rm m/s
$$

Angular velocity is:

$$
omega=frac{5}{0.3}
$$

$$
boxed{omega=16.67,,rm rad/s}
$$

Lets calculate torque:

$$
tau=40cdot0.3=12,,rm N
$$

Moment of inertia is:

$$
I=frac{12}{8.33}
$$

$$
boxed{I=1.44,,rm kgm^2}
$$

$$
alpha=8.33,,rm rad/s^2
$$

$$
omega=16.67,,rm rad/s
$$

$$
I=1.44,,rm kgm^2
$$