Respuesta :
Answer:
A. pi1 = 690 kg m/s
B. Δp1 = - 690 kg m/s
C. The two changes in momenta are equal in magnitude and opposite in direction.
D. v2i = -6.16 m/s
Explanation:
Part A.
The momentum can be calculated as mass times velocity, so the initial momentum of the 92-kg player is equal to
pi = mv
pi = (92 kg)(7.5 m/s)
pi = 690 kg m/s
So, the initial momentum of the 92-kg player was 690 kg m/s
Part B.
After the collision, each player has zero velocity, so the final velocity will be 0 m/s. Then, the change in momentum is equal to
[tex]\begin{gathered} \Delta p_1=p_f-p_i=mv_f-p_i_{} \\ \Delta p_1=92\operatorname{kg}(0\text{ m/s) - }690\operatorname{kg}\text{ m/s} \\ \Delta p_1=0\operatorname{kg}\text{ m/s - 690 kg m/s} \\ \Delta p1=-690\text{ kg m/s} \end{gathered}[/tex]Part C.
In a closed system the momentum is conserved, so to conserve the momentum the change in momentum of the 92-kg player should be equal in magnitude but opposite in direction. Then, when we add them, the sum should be equal to 0. So, the answer is:
The two changes in momenta are equal in magnitude and opposite in direction.
Part D.
Taking into account part C, we can write the following equation
[tex]\begin{gathered} \Delta p_2=-\Delta p_1 \\ m_2v_f-m_2v_{i2}=-\Delta p_1 \end{gathered}[/tex]Since the final velocity of the 112-kg player is also 0 m/s, we can solve for v2i as
[tex]\begin{gathered} m_2(0)-m_2v_{2i}=-\Delta p_1 \\ -m_2v_{2i}=-\Delta p_1 \\ v_{2i}=\frac{-\Delta p_1}{-m_2}=\frac{\Delta p_1}{m_2} \end{gathered}[/tex]Replacing the values, we get
[tex]v_{2i}=\frac{-690\text{ kg m/s}}{112\operatorname{kg}}=-6.16\text{ m/s}[/tex]Therefore, the answers are:
A. pi1 = 690 kg m/s
B. Δp1 = - 690 kg m/s
C. The two changes in momenta are equal in magnitude and opposite in direction.
D. v2i = -6.16 m/s
