Question

Two people are initially at rest and touching each other. Theypush off each other and they move away in opposite directions.Person 1 has a mass of 87.5 kg and a velocity of -10 meters persecond. Person 2 has a mass of 39.2 kg. What is the velocity ofPerson 2?V =unitHow far apart will the two people be after 3 seconds? (Hint,both are moving away. Calculate the distance each will traveland add those distances, or get the net velocity and calculatethe distance)d =unit

Answer 1

Answer:

$\begin{gathered} v_2=22.321ms^{-1} \\ d=96.964m\Rightarrow\text{ ( Distance apart )} \end{gathered}$

Explanation: This problem can be solved using the conservation of momentum concept, the equations used are as follows:

$\begin{gathered} p_i=p_f \\ p_i=p_1+p_2=0\Rightarrow p_i=(m_1+m_2)v_i=0\Rightarrow(1) \\ p_f=p_1+p_2=0\Rightarrow p_f=m_1v_1+m_2v_2=0\Rightarrow(2) \end{gathered}$

The knowns and unknowns in equation (1) and (2) are as follows:

$\begin{gathered} m_1=87.5\operatorname{kg} \\ v_1=-10ms^{-1} \\ m_2=39.2\operatorname{kg} \\ v_2=\text{?} \\ v_i=0 \end{gathered}$

Plugging these values in (1) and (2) gives us the following result:

$\begin{gathered} (1)=(2) \\ \therefore\Rightarrow \\ (m_1+m_2)(0)=m_1v_1+m_2v_2=0 \\ (87.5\operatorname{kg})(-10ms^{-1})+(39.2\operatorname{kg})v_2=0 \\ v_2=\frac{(87.5\operatorname{kg})(10ms^{-1})}{(39.2\operatorname{kg})}=22.321ms^{-1} \\ v_2=22.321ms^{-1} \end{gathered}$

To calculate the distance travelled in 3 seconds, we simply do as follows:

$\begin{gathered} s=vt \\ \therefore\Rightarrow \\ (1)\rightarrow s_1=(-10ms^{-1})\cdot(3s)=-30m \\ (2)\rightarrow s_2=(22.321ms^{-1})(3s)=66.964m \\ \therefore\Rightarrow \\ d=|s_1|+|s_2|=|-30m|+|66.964m|=96.964m \\ d=96.964m \end{gathered}$