(c) find the change in kinetic energy of the system.
calculate the initial kinetic energy of the system.
$$
\begin{aligned}
ke_{i}=& \frac{1}{2} m_{1} v_{1 i}^{2}+\frac{1}{2} m_{2} v_{2 i}^{2} \\
=& \frac{1}{2}\left(1.80 \times 10^{3} \mathrm{~kg}\
ight)(15.0 \mathrm{~m} / \mathrm{s})^{2} \\
&+\frac{1}{2}\left(9.00 \times 10^{2} \mathrm{~kg}\
ight)(-15.0 \mathrm{~m} / \mathrm{s})^{2} \\
=& 3.04 \times 10^{5} \mathrm{~j}
\end{aligned}
$$
calculate the final kinetic energy of the system and the change in kinetic energy, $\delta k e$.
$$
\begin{aligned}
k e_{f}=& \frac{1}{2}\left(m_{1}+m_{2}\
ight) v_{f}^{2} \\
=& \frac{1}{2}\left(1.80 \times 10^{3} \mathrm{~kg}+9.00 \times 10^{2} \mathrm{~kg}\
ight)(5.00 \mathrm{~m} / \mathrm{s})^{2} \\
=& 3.38 \times 10^{4} \mathrm{~j}
\end{aligned}
$$
$$
\delta k e=k e_{f}-k e_{i}=-2.70 \times 10^{5} \mathrm{~j}
$$
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remarks during the collision, the system lost almost $90 \%$ of its kinetic energy. the change in velocity of the pickup truck was only $10.0 \mathrm{~m} / \mathrm{s}$, compared to twice that for the compact car. this example underscores perhaps the most important safety feature of any car: its mass. injury is caused by a change in velocity, and the more massive vehicle undergoes a smaller velocity change in a typical accident.
question if the mass of both vehicles were doubled, how would the final velocity and the change in kinetic energy be affected? (select all that apply.)
$ \square $ the change in kinetic energy would be half as great.
$ \square $ the change in kinetic energy would be 2 times as great.
$ \square $ the final velocities would each be $1 / 2$ times as large in magnitude.
$ \square $ the final velocities would each be 2 times as large in magnitude.
$ \square $ the final velocities would each have the same magnitude as before.
$ \square $ the change in kinetic energy would be unchanged.

Question

Accepted Answer

# Explanation:
## Step1: Calculate final velocity using conservation of momentum
According to the law of conservation of momentum $m_1v_{1i}+m_2v_{2i}=(m_1 + m_2)v_f$. If $m_1' = 2m_1$ and $m_2'=2m_2$, then $2m_1v_{1i}+2m_2v_{2i}=(2m_1 + 2m_2)v_f'$. Factor out 2 on both sides: $2(m_1v_{1i}+m_2v_{2i})=2(m_1 + m_2)v_f'$. Divide both sides by 2, we get $m_1v_{1i}+m_2v_{2i}=(m_1 + m_2)v_f'$. So $v_f'=v_f$ (the final velocity remains the same).
## Step2: Calculate change in kinetic energy
Initial kinetic energy $KE_i=\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2$. New initial kinetic energy $KE_i'=\frac{1}{2}(2m_1)v_{1i}^2+\frac{1}{2}(2m_2)v_{2i}^2 = 2(\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2)=2KE_i$. Final kinetic energy $KE_f=\frac{1}{2}(m_1 + m_2)v_f^2$. New final kinetic energy $KE_f'=\frac{1}{2}(2m_1 + 2m_2)v_f^2=2(\frac{1}{2}(m_1 + m_2)v_f^2)=2KE_f$. Change in kinetic energy $\Delta KE=KE_f - KE_i$, new change in kinetic energy $\Delta KE'=KE_f' - KE_i'=2(KE_f - KE_i)=2\Delta KE$

# Answer:
The change in kinetic energy would be 2 times as great. The final velocities would each have the same magnitude as before.

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate final velocity using conservation of momentum

Step2: Calculate change in kinetic energy

Answer: