Question

Determine the velocities of two masses after collision of a 6 – kg mass moving with a velocity of 6 m/s and a 4 – kg mass that is moving with a velocity of 3 m/s if there are no external forces acting on them.

A. e=Vs/Va

B. What are the given data?

C-1. Calculate using the law on conservation of momentum: (equation 1)

mAuA+mBuB=mAVA+mBVB

C-2. and the coefficient of restitution: (equation 2)

e=VB-VA/uB-uA

D. Substitute in equation 1:

E. Substitute in equation 2:

F. Conclusion:

Answer 1

$v_1=\frac{m_1-m_2}{m_1+m_2}\cdot u_1+\frac{2m_2}{m_1+m_2}\cdot u_2=$

$=\frac{6-4}{6+4}\cdot6+\frac{2\cdot 4}{6+4}\cdot 3=3.6\ (m/s)$

$v_2=\frac{m_2-m_1}{m_1+m_2}\cdot u_2+\frac{2m_1}{m_1+m_2}\cdot u_1=$

$=\frac{4-6}{6+4}\cdot 3+\frac{2\cdot 6}{6+4}\cdot 6=6.6\ (m/s)$

The coefficient of restitution

$e=\frac{v_2-v_1}{u_1-u_2}=\frac{6.6-3.6}{6-4}=1.5$