Respuesta :
Answer:
The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.
Explanation:
Given that,
Mass of the cart, m = 250 g = 0.25 kg
Initial velocity of the cart, u = 0.31 m/s (due right)
Mass of another cart, m' = 500 g = 0.5 kg
Initial velocity of the another cart u' = -0.22 m/s (due left)
Let p is the total momentum of the system before the collision. It is given by :
[tex]p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s[/tex]
So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.
Answer:
Explanation:
mass of I cart, m = 250 g = 0.25 kg
initial velocity of I cart, u = 0.31 m/s
mass of II cart, m' = 500 g =0.5 kg
initial velocity of II cart, u' = - 0.22 m/s
Total momentum before collision = m x u + m' x u'
= 0.25 x 0.31 - 0.5 x 0.22
= 0.0775 - 0.11
= - 0.0325 kg m/s
