All Solutions
Page 233: Lesson Check
Solve for the first case $m_{1} = m$:
In order to evaluate the magnitude of the Momentum, we use the formula that is given in the second yellow box:
$$
begin{align*}
p_{1} &= m_{1} ~ v \
&= m ~ v
end{align*}
$$
Solve for the second case $m_{2} =2 m$:
In order to evaluate the magnitude of the Momentum, we use the formula that is given in the second yellow box:
$$
begin{align*}
p_{2} &= m_{2} ~ v \
&=2 m ~ v \
&= 2 p_{1}
end{align*}
$$
So, in the case of doubling the mass of the object, the momentum will double, too.
In order to evaluate the Momentum vector, we use the formula that is given in the first yellow box:
$$
begin{align*}
vec{p} &= vec{mv} \
&= m ~ vec{v} \
end{align*}
$$
So, the momentum vector is the product of the mass which is a scalar, and the velocity which is a vector. Therefore, the direction of the momentum vector is only depending on the direction of the velocity in the same direction.
In order to evaluate the magnitude of the Momentum, we use the formula that is given in the second yellow box:
$$
begin{align*}
p &= m ~ v \
end{align*}
$$
So, in the case of doubling the mass of the object, the momentum will double, too.
Solve for the case that the speed of the object is $v_{1} = v$:
In order to evaluate the kinetic energy of the object, we use the following relation:
$$
begin{align*}
K.E_{1} &= dfrac{1}{2} ~ m ~ v_{1}^{2} \
&= dfrac{1}{2} ~ m ~ v^{2} \
end{align*}
$$
Solve for the case that the speed of the object is $v_{2} = 2 v$:
By substituting by the double speed of the object, then we get
$$
begin{align*}
K.E_{2} &= dfrac{1}{2} ~ m ~ v^{2} \
&= dfrac{1}{2} ~ m ~ left( 2 v right)^{2} \
&= dfrac{1}{2} ~ m ~ 4 ~ v^{2} \
&= 4 ~ dfrac{1}{2} ~ m ~ v^{2} \
&= 4 K.E_{1}
end{align*}
$$
So, in the case of doubling the mass of the object, the kinetic energy will increase four times.
$$
KE=frac{1}{2}m(2v)^2=4timesfrac{1}{2}mv^2
$$
In order to evaluate the Momentum vector, we use the formula that is given in the first yellow box:
$$
begin{align*}
vec{p} &= vec{mv} \
&= m ~ vec{v} \
end{align*}
$$
So, the total momentum vector of the two particles is equal to
$$
begin{align*}
vec{p_{T}} &= vec{p_{1}} + vec{p_{2}} \
&= m_{1} ~ vec{v_{1}} + m_{2} ~ vec{v_{2}} \
&= vec{0} \
end{align*}
$$
So, the zero momentum is a zero vector which means that either the sum magnitude of the two momentums is zero or the direction of the sum magnitude of the two momentums is zero.
In order to evaluate the kinetic energy, we use the following relation:
$$
begin{align*}
K.E &= dfrac{1}{2} ~ m ~ v^{2} \
end{align*}
$$
So, the total kinetic energy of the two particles, which is a scalar, is equal to
$$
begin{align*}
K.E_{T} &= K.E_{1} + K.E_{2} \
&= dfrac{1}{2} ~ m_{1} ~ v_{1}^{2} + dfrac{1}{2} m_{2} ~ v_{2}^{2} \
end{align*}
$$
Since the zero momentum is a zero vector which means that either the summed magnitude of the two momentums is zero or the direction of the summed magnitude of the two momentums is zero. We have two cases, if the sum of the square of the two speeds is zero, then the total kinetic energy is zero. So, the counterexample, if the sum of the square of the two speeds isn’t zero and the two velocities are in opposite directions, then the total kinetic energy isn’t zero. Because the kinetic energy is a scalar (doesn’t depend on the direction).
Product of mass and speed will give us momentum.
$textbf{Solution:}$ Multiply mass with speed to get $p$
$$
p=mv=(0.16,kg)(2.7,m/s)=0.43,kg.m/s
$$
$$
boxed{bf color{#4257b2}p=0.43,kg.m/s}
$$
p=0.43,kg.m/s
$$
The mass of the school’s bus is $m_{s} = 1.82 times 10^{4} mathrm{~kg}$. The speed of the school’s bus is $v_{s} = 12.5 mathrm{~m/s}$. The mass of the baseball is $m_{b} = 0.142 mathrm{~kg}$.
$textbf{Required: }$
Finding the speed of the baseball, if the momentum of the school’s bus equals the momentum of the baseball.
Solve for the school’s bus:
In order to evaluate the magnitude of the Momentum of the school’s bus, we use the formula that is given in the second yellow box:
$$
begin{align*}
p_{s} &= m_{s} ~ v_{s} \
&= 1.82 times 10^{4} mathrm{~kg} times 12.5 mathrm{~m/s} \
&= 2.275 times 10^{5} mathrm{~kg cdot m/s} \
end{align*}
$$
Solve for the baseball:
In order to evaluate the magnitude of the Momentum of the baseball, we use the formula that is given in the second yellow box:
$$
begin{align*}
p_{b} &= m_{b} ~ v_{b} \
end{align*}
$$
Rearrange and solve for the speed of the baseball:
$$
begin{align*}
v_{b} &= dfrac{ p_{b} }{ m_{b} } \
&= dfrac{ 2.275 times 10^{5} mathrm{~kg cdot m/s} }{ 0.142 mathrm{~kg} } \
&= 1.602 times 10^{6} mathrm{~m/s}
end{align*}
$$
So, the speed of the baseball is $1.602 times 10^{6} mathrm{~m/s}$.
The speed of the running dog is $v = 1.9 mathrm{~m/s}$. The momentum of the running dog is $p = 17 mathrm{~kg cdot m/s}$.
$textbf{Required: }$
Finding the mass of the dog.
In order to evaluate the magnitude of the Momentum of the running dog, we use the formula that is given in the second yellow box:
$$
begin{align*}
p &= m ~ v \
end{align*}
$$
Rearrange and solve for the dog’s mass:
$$
begin{align*}
m &= dfrac{ p }{ v } \
&= dfrac{ 17 mathrm{~kg cdot m/s} }{1.9 mathrm{~m/s} } \
&= 8.947 mathrm{~kg}
end{align*}
$$
So, the dog’s mass is $8.947 mathrm{~kg}$.
The mass of the cart $1$ is $m_{1} = 0.35 mathrm{~kg}$. The speed of the cart $1$ is $v_{1} = 1.2 mathrm{~m/s}$. The mass of the cart $2$ is $m_{2} = 0.61 mathrm{~kg}$. The speed of the cart $2$ is $v_{2} = 0.85 mathrm{~m/s}$.
$textbf{Required: }$
Finding the total momentum of the system.
In order to evaluate the Momentum vector, we use the formula that is given in the first yellow box:
$$
begin{align*}
vec{p} &= vec{mv} \
&= m ~ vec{v} \
end{align*}
$$
Let $hat{e}$ is the direction of cart $1$ motion. So, the total momentum vector of the two carts is equal to
$$
begin{align*}
vec{p_{T}} &= vec{p_{1}} + vec{p_{2}} \
&= m_{1} ~ vec{v_{1}} + m_{2} ~ vec{v_{2}} \
&= 0.35 mathrm{~kg} times 1.2 mathrm{~m/s} ~ hat{e} + 0.61 mathrm{~kg} times 0.85 mathrm{~m/s} ~ left( -hat{e} right) \
&= 0.35 mathrm{~kg} times 1.2 mathrm{~m/s} ~ hat{e} – 0.61 mathrm{~kg} times 0.85 mathrm{~m/s} ~hat{e} \
&= – left(0.0985 mathrm{~kg cdot m/s} right)~ hat{e}
end{align*}
$$
So, the total momentum of the system is $0.0985 mathrm{~kg cdot m/s}$ in the opposite direction of the cart $1$ motion.
