All Solutions
Page 252: Practice Problems
The mass of the touchdown is $m_{1} = 95 mathrm{~kg}$. The initial speed of the touchdown is $v_{i,1} = 3.75 mathrm{~m/s}$. The mass of the linebacker is $m_{2} = 111 mathrm{~kg}$. The initial speed of the linebacker is $v_{i,2} = 4.1 mathrm{~m/s}$. The final speed of the two players is equal to zero.
$textbf{Required: }$
Finding if the running back runs faster or slower.
According to the Momentum-Impulse Theorem:
$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
&= left( m ~ vec{v_{f}} – m ~ vec{v_{i}} right) \
&= m ~ left( vec{v_{f}} – vec{v_{i}} right) \
end{align*}
$$
According to conservation law of momentum, the impulse is equal to zero.
$$
begin{align*}
Delta p_{t} &= Delta p_{1} + Delta p_{2} \
&= m_{1} ~ left( vec{v_{f,1}} – vec{v_{i,1}} right) + m_{2} ~ left( vec{v_{f,2}} – vec{v_{i,2}} right) \
&= m_{1} ~ left( 0 – vec{v_{i,1}} right) + m_{2} ~ left( 0 – vec{v_{i,2}} right) \
&= 95 mathrm{~kg} times left( – 3.75 mathrm{~m/s} right) + 111 mathrm{~kg} times left( – 4.1 mathrm{~m/s} right) \
&= – 811.35 mathrm{~kg cdot m/s}
end{align*}
$$
From example $7.10$, the immediate velocity after the collision is $- 0.48 mathrm{~m/s}$ to the left.
According to the conservation law of momentum assigns that the total momentum of the system is equal to zero. if they want to run backward, they will need more momentum to nullify the effect of left-sided momentum, and only this way the system momentum will be equal to zero.
The overall momentum of the two players is towards the left, showing that the linebacker has greater momentum towards the left than the running back’s momentum towards the right. We found that running back should run faster to reach increased momentum towards the right to make the overall momentum of the system zero.
The mass of the touchdown is $m_{1} = 95 mathrm{~kg}$. The initial speed of the touchdown is $v_{i,1} = 3.75 mathrm{~m/s}$. The mass of the linebacker is $m_{2} = 111 mathrm{~kg}$. The initial speed of the linebacker is $v_{i,2} = 4.1 mathrm{~m/s}$.
$textbf{Required: }$
Finding final speed for the running back player.
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 final speed for the running back player:
$$
begin{align*}
v_{f,1} &= – dfrac{ m_{2} ~ v_{f,2} }{ m_{1} } \
&= – dfrac{ 111 mathrm{~kg} times left( – 4.1 mathrm{~m/s} right) }{95 mathrm{~kg} } \
&= 4.791 mathrm{~m/s}
end{align*}
$$
So, the final speed for the running back player is $4.791 mathrm{~m/s}$.
The mass of the wooden block is $m_{1} = 1.35 mathrm{~kg}$. The initial speed of the wooden block is $v_{i,1} = 0 mathrm{~m/s}$. The mass of the bullet is $m_{2} = 0.0105 mathrm{~kg}$. The initial speed of the bullet is $v_{i,2} = 715 mathrm{~m/s}$.
$textbf{Required: }$
Finding final speed of the bullet-block system after the collision.
According to the Momentum-Impulse Theorem:
$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
&= left( m ~ vec{v_{f}} – m ~ vec{v_{i}} right) \
&= m ~ left( vec{v_{f}} – vec{v_{i}} right) \
end{align*}
$$
According to the conservation law of momentum, the impulse is equal to zero.
$$
begin{align*}
Delta p_{t} &= Delta p_{1} + Delta p_{2} \
&= 0 \
&= m_{1} ~ left( vec{v_{f,1}} – vec{v_{i,1}} right) + m_{2} ~ left( vec{v_{f,2}} – vec{v_{i,2}} right) \
&= m_{1} ~ left( vec{v_{f,1}} – v_{i,1} right) + m_{2} ~ left( vec{v_{f,2}} – v_{i,2} right) \
&= m_{1} ~ left( vec{v_{f,1}} – 0 right) + m_{2} ~ left( vec{v_{f,2}} – v_{i,2} right) \
&= left( m_{1} + m_{2} right) ~ v_{f} + m_{2} ~ v_{i,2} \
end{align*}
$$
Rearrange and solve for the final speed of the bullet-block system after the collision:
$$
begin{align*}
v_{f} &= – dfrac{ m_{2} ~ v_{f,1} }{ left( m_{1} + m_{2} right) } \
&= – dfrac{ 0.0105 mathrm{~kg} times left( – 715 mathrm{~m/s} right) }{ left( 1.35 mathrm{~kg} + 0.0105 mathrm{~kg} right) } \
&= 5.518 mathrm{~m/s}
end{align*}
$$
So, the final speed of the bullet-block system after the collision is $5.518 mathrm{~m/s}$.
The mass of the car is $m_{1} = 1200 mathrm{~kg}$. The initial speed of the car is $v_{i,1} = 2.5 mathrm{~m/s}$. The mass of the truck is $m_{2} = 2600 mathrm{~kg}$. The initial speed of the truck is $v_{i,2} = 6.2 mathrm{~m/s}$. The two vehicles stick together after the collision. Neglect the external forces.
$textbf{Required: }$
Finding the immediate speed after the collision.
According to the Momentum-Impulse Theorem:
$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
&= left( m ~ vec{v_{f}} – m ~ vec{v_{i}} right) \
&= m ~ left( vec{v_{f}} – vec{v_{i}} right) \
end{align*}
$$
According to the conservation law of momentum, the impulse is equal to zero.
$$
begin{align*}
Delta p_{t} &= Delta p_{1} + Delta p_{2} \
&= 0 \
&= m_{1} ~ left( vec{v_{f,1}} – vec{v_{i,1}} right) + m_{2} ~ left( vec{v_{f,2}} – vec{v_{i,2}} right) \
&= m_{1} ~ left( vec{v_{f,1}} – v_{i,1} right) + m_{2} ~ left( vec{v_{f,2}} – v_{i,2} right) \
&= m_{1} ~ left( vec{v_{f,1}} – v_{i,1} right) + m_{2} ~ left( vec{v_{f,2}} – v_{i,2} right) \
&= left( m_{1} + m_{2} right) ~ v_{f} – m_{1} ~ v_{i,1} – m_{2} ~ v_{i,2} \
end{align*}
$$
Rearrange and solve for the final speed of the bullet-block system after the collision:
$$
begin{align*}
v_{f} &= dfrac{ m_{1} ~ v_{i,1} + m_{2} ~ v_{i,2} }{ left( m_{1} + m_{2} right) } \
&= dfrac{ 1200 mathrm{~kg} times left( 2.5 mathrm{~m/s} right) + 2600 mathrm{~kg} times left( 6.2 mathrm{~m/s} right) }{ left( 1200 mathrm{~kg} + 2600 mathrm{~kg} right) } \
&= 5.032 mathrm{~m/s}
end{align*}
$$
So, the immediate speed after the collision is $5.032 mathrm{~m/s}$.
Haven't found what you were looking for?
Search for samples, answers to your questions and flashcards
