Question

now jed and kadia tackle a homework problem:
an object of mass ( m_1 = 20 mathrm{~kg} ) and velocity ( vec{v}_1 = 6.5 mathrm{~m} / mathrm{s} ) crashes into another object of mass ( m_2 = 5 mathrm{~kg} ) and velocity ( vec{v}_2 = -16.5 mathrm{~m} / mathrm{s} ). the two particles stick together as a result of the collision. because no external forces are acting, the collision does not change the total momentum of the system of two particles, so the principle of conservation of linear momentum applies.
m_1 v_{1 i}+m_2 v_{2 i}=left(m_1+m_2
ight) v_f
if jed and kadia use the one - dimensional conservation of momentum equation to find the final velocity (in ( mathrm{m} / mathrm{s} )) of the two joined objects after the collision, what do they obtain? (indicate the direction with the sign of your answer.)
( mathrm{m} / mathrm{s} )

Explanation:

Step1: Substitute values into formula

Given \(m_1 = 20\space kg\), \(v_{1i}=6.5\space m/s\), \(m_2 = 5\space kg\), \(v_{2i}=- 16.5\space m/s\) into \(m_1v_{1i}+m_2v_{2i}=(m_1 + m_2)v_f\).
So, \(20\times6.5+5\times(-16.5)=(20 + 5)v_f\).

Step2: Calculate left - hand side

First, \(20\times6.5 = 130\) and \(5\times(-16.5)=-82.5\).
Then \(130-82.5=(20 + 5)v_f\), i.e., \(47.5 = 25v_f\).

Step3: Solve for \(v_f\)

\(v_f=\frac{47.5}{25}\).

Answer:

\(1.9\space m/s\)