Question

6.If a 250. gram cart moving to the right with a velocity of .31 m/s collides inelastically with a 500. gram cart traveling to the left with a velocity of -0.22 m/s, what is the total momentum of the system before the collision

Answer 1

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 :

$p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s$

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.            

Answer 2

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