All Solutions
Page 241: Lesson Check
As the textbook assigns that the impulse is the product of a force and the time which is given by
$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
end{align*}
$$
According to the Momentum-Impulse Theorem:
$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
&= left( m ~ vec{v_{f}} – m ~ vec{v_{i}} right) \
&= m ~ left( vec{v_{f}} – vec{v_{i}} right) \
end{align*}
$$
So, the impulse vector has the same direction as the force as in first equation. Also, the impulse vector has the same direction as the changing of the momentum but isn’t the same as the momentum. As we know the kinetic energy is a scalar quantity.
As the textbook assigns that the impulse is the product of a force and the time which is given by
$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
end{align*}
$$
According to the Momentum-Impulse Theorem:
$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
&= left( m ~ vec{v_{f}} – m ~ vec{v_{i}} right) \
&= m ~ left( vec{v_{f}} – vec{v_{i}} right) \
end{align*}
$$
So, the impulse vector has the same direction as the force as in first equation. Also, the impulse vector has the same direction as the changing of the momentum. Also, the value of the impulse is given by the second equation which assigns that the impulse is given by changing of the momentum vector. But the impulse doesn’t depend directly on the momentum.
As the textbook assigns that the impulse is the product of a force and the time which is given by
$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
end{align*}
$$
According to the Momentum-Impulse Theorem:
$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
&= left( m ~ vec{v_{f}} – m ~ vec{v_{i}} right) \
&= m ~ left( vec{v_{f}} – vec{v_{i}} right) \
end{align*}
$$
So, the impulse vector is depending of the acting force and the small time interval of the accident as in first equation. Also, the impulse vector has the same direction as the changing of the momentum. Also, the value of the impulse is given by the second equation which assigns that the impulse is given by changing of the momentum vector. But the impulse doesn’t depend directly on the momentum. But the main factor of the impulse is the time of the accident. So, the steel helmet elongate the time of the accident. So, as in first equation, if the time interval of the accident increases, then the helmet will absorb the accident. So, the helmet minimize the time interval of the collision between the helmet and the head of the victim. Therefore, the impulse will decreases.
Product of force and duration for each case will give impulse, and then we can compare them.
$textbf{Solution:}$ Calculating the impulse gives
$$
I_A=F_ADelta t_A = (F)(Delta t) = F Delta t
$$
$$
I_B=F_BDelta t_B = (2F)(Delta t/3) = frac{2}{3}F Delta t
$$
$$
I_C=F_CDelta t_C = (5F)(Delta t/10) = frac{1}{2}F Delta t
$$
$$
I_D=F_DDelta t_D = (10F)(Delta t/100) = frac{1}{10}F Delta t
$$
Comparing the impulses gives us the ranking
$$
boxed{bf color{#4257b2} {I_D<I_C=I_B<I_A}}
$$
I_D<I_C=I_B<I_A
$$
For the case $1$, the car hits a light pole and comes to rest. For the case $2$, the car hits a pile of plastic garbage bags and comes to rest.
Since the mass of the car is still the same in both cases. Also, the initial velocity of the car is the same in both cases $vec{v_{i,1}} = vec{v_{i,2}}$. The car comes to rest in both cases $vec{v_{f,1}} = vec{v_{f,2}}$.
Solve for the case $1$:
According to the Momentum-Impulse Theorem:
$$
begin{align*}
vec{I_{1}} &= vec{F_{1}} ~ Delta t \
&= Delta vec{p,1} \
&= left( vec{p_{f,1}} – vec{p_{i,1}} right) \
&= left( m ~ vec{v_{f,1}} – m ~ vec{v_{i,1}} right) \
&= m ~ left( vec{v_{f,1}} – vec{v_{i,1}} right) \
end{align*}
$$
Solve for the case $2$:
According to the Momentum-Impulse Theorem:
$$
begin{align*}
vec{I_{2}} &= vec{F_{2}} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f_{2}}} – vec{p_{i_{2}}} right) \
&= left( m ~ vec{v_{f_{2}}} – m ~ vec{v_{i_{2}}} right) \
&= m ~ left( vec{v_{f_{2}}} – vec{v_{i_{2}}} right) \
&= m ~ left( vec{v_{f_{1}}} – vec{v_{i_{1}}} right) \
&= vec{I_{1}}
end{align*}
$$
So, the impulse in a case $1$ is equal to the impulse in a case $2$.
Since the mass of the car is still the same in both cases. Also, the initial velocity of the car is the same in both cases $vec{v_{i,1}} = vec{v_{i,2}}$. The car comes to rest in both cases $vec{v_{f,1}} = vec{v_{f,2}}$. the impulse in case $1$ is equal to the impulse in case $2$.
Solve for the case $1$:
As the textbook assigns that the impulse is the product of a force and the time which is given by
$$
begin{align*}
vec{I_{1}} &= vec{F_{1}} ~ Delta t_{1} \
end{align*}
$$
Solve for the case $2$:
As the textbook assigns that the impulse is the product of a force and the time which is given by
$$
begin{align*}
vec{I_{2}} &= vec{F_{2}} ~ Delta t_{2} \
end{align*}
$$
Divide both equation together, then we get
$$
begin{align*}
dfrac{ vec{I_{2}} }{ vec{I_{1}} } &= dfrac{ vec{F_{2}} ~ Delta t_{2} }{ vec{F_{1}} ~ Delta t_{1} } \
&= dfrac{ vec{I_{1}} }{ vec{I_{1}} } \
&= 1
end{align*}
$$
Rearrange and solve for the average force in the case $2$:
$$
begin{align*}
vec{F_{2}} &= vec{F_{1}} ~ dfrac{ Delta t_{1} }{ Delta t_{2} } \
end{align*}
$$
In case $2$, the car hits a pile of plastic garbage bags. So, the reaction force by the pile of plastic garbage bags on the car is less than the reaction force by the light pole on the car. So, the time interval in case $2$ is greater than the time interval in case $1$ ~ $Delta t_{2} > Delta t_{1}$. Substituting in the previous equation, then the average force in a case $1$ is greater than the average force in a case $2$.
(b) The average force in case $1$ is greater than the average force in case $2$.
The time interval in the first case is $Delta t_{1} = 0.005 mathrm{~s}$. The time interval in the second case is $Delta t_{2} = 0.02 mathrm{~s}$.
$textbf{Required: }$
Finding the ratio between the two forces.
Solve for the case $1$:
As the textbook assigns that the impulse is the product of a force and the time which is given by
$$
begin{align*}
vec{I_{1}} &= vec{F_{1}} ~ Delta t_{1} \
end{align*}
$$
Solve for the case $2$:
As the textbook assigns that the impulse is the product of a force and the time which is given by
$$
begin{align*}
vec{I_{2}} &= vec{F_{2}} ~ Delta t_{2} \
end{align*}
$$
Divide both equation together, then we get
$$
begin{align*}
dfrac{ vec{I_{2}} }{ vec{I_{1}} } &= dfrac{ vec{F_{2}} ~ Delta t_{2} }{ vec{F_{1}} ~ Delta t_{1} } \
end{align*}
$$
Rearrange and solve for the ratio between the two forces:
$$
begin{align*}
dfrac{ vec{F_{2}} }{ vec{F_{1}} } &= dfrac{ vec{I_{2}} }{ vec{I_{1}} } ~ dfrac{ Delta t_{1} }{ Delta t_{2} } \
&= dfrac{ Delta vec{p_{2}} }{ Delta vec{p_{1}} } ~ dfrac{ Delta t_{1} }{ Delta t_{2} } \
end{align*}
$$
Since the two changing in the momentum in two cases is the same, Then we get
$$
begin{align*}
dfrac{ vec{F_{2}} }{ vec{F_{1}} } &= dfrac{ Delta vec{p_{2}} }{ Delta vec{p_{1}} } ~ dfrac{ Delta t_{1} }{ Delta t_{2} } \
&= dfrac{ Delta vec{p_{1}} }{ Delta vec{p_{1}} } ~ dfrac{ Delta t_{1} }{ Delta t_{2} } \
&= dfrac{ Delta t_{1} }{ Delta t_{2} } \
&= dfrac{ 0.005 mathrm{~s} }{ 0.02 mathrm{~s} } \
&= dfrac{1}{4}
end{align*}
$$
So, the average force is reduced by $4$, ~ $vec{F_{2}} = dfrac{1}{4} ~ vec{F_{1}}$.
The time interval that the ball takes in contact is $Delta t = 0.001 mathrm{~s}$. The mass of the ball is $m = 0.045 mathrm{~kg}$. The final speed of the ball is $v_{f} = 67 mathrm{~m/s}$.
$textbf{Required: }$
Finding the magnitude of the exerted force on the ball by the club.
According to the Momentum-Impulse Theorem:
$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
&= left( m ~ vec{v_{f}} – m ~ vec{v_{i}} right) \
&= m ~ left( vec{v_{f}} – vec{v_{i}} right) \
end{align*}
$$
Rearrange and solve for the magnitude of the exerted force on the ball by the club:
$$
begin{align*}
F &= dfrac{ I }{ Delta t } \
&= dfrac{ m ~ left( v_{f} – v_{i} right) }{ Delta t } \
&= dfrac{ 0.045 mathrm{~kg} times left( 67 mathrm{~m/s} – 0 mathrm{~m/s} right) }{ 0.001 mathrm{~s} } \
&= 3.015 times 10^{3} mathrm{~N}
end{align*}
$$
So, the magnitude of the exerted force on the ball by the club is $3.015 times 10^{3} mathrm{~N}$.
$$
F_{av} = dfrac{Delta p}{Delta t}
$$
where $Delta p$ can be exressed as
$$
Delta p = p_{f} – p_{i} = mv_{f} – mv_{i}
$$
$$
Delta t = dfrac{mv_{f} – mv_{i}}{F_{av}} = dfrac{(3.2m/s)(0.50kg) – (0)(0.50kg)}{230N}
$$
$$
Delta t = 0.007s
$$
textbf{Concept:}
$$
Change in the momentum of teh croquet ball is to be found and then $Delta t$ is to be found using momentum-impulse theorem
$$
textbf{Solution:}
$$
$$
Delta t=frac{Delta p}{F_{av}}=frac{m(v_f-v_i)}{F_{av}}=frac{(0.50,kg)(3.2,m/s-0)}{230,N}
$$
$$
boxed{bf color{#4257b2}=0.007,s}
$$
7,ms;or;0.007;s
$$
