What is the momentum of the cart?

Question

A cart with mass 200. g has a velocity of 0.30 m/s in the positive x direction. What is the momentum of the cart?
This cart collides with a stationary cart of mass 400 g.
If the velocity of the first cart is -0.1000 m/s after the collision, what is the velocity of the second cart after the collision?What type of collision is this?

Answer 1

Momentum p=mV
Momentum of the cart is
p= 0.200 x 0.30 = 0.06 kg m/s

Conservation of momentum

m1u1+m2u2=m1v1 + m2v2
in this case velocity of cart 2 is u2=0

m1u1=m1v1 + m2v2 then

v2=(m1u1 - m1v1 )/m2
v2=(m1/m2)( u1 - v1)
v2=(.200/.400)( 0.30 - (-0.100))=0.20 m/s

Collision is perfectly elastic. If it was inelastic there would have been loss in momentum into energy like heat.

Answer 2

Mass of first cart = m1 =200g=0.2 kg

Velocity of the first cart before collision = u1=0.30 m/s

momentum of first cart before collision p1=0.2*0.3=0.06 kgm/s

Mass of second cart = m2 =400g=0.4 kg

Velocity of the second cart before collision = u2=zero

momentum of second cart before collision p2= zero

the velocity of the first cart after the collision v1= -0.1000 m/s

momentum of the first cart ,after the collision pf1= -0.02 kgm/s

If v2 is velocity of second cart after collision,

momentum of the second cart ,after the collision pf2= 0.04 v kgm/s

As momentum is always conserved in a collision,

pf1+pf2=p1+p2

- 0.02 +0.4 v= 0.06 + zero

0.4 v=0.08

v= 0.2 m/s

the velocity of the second cart after the collision is 0.2 m/s in + x direction

As initial kinetic energy of system is 0.009 J and final kinetic of system is 0.009 , kinetic energy is conserved, hence collision is ELASTIC