From the knowledge of linear momentum,
a.) momentum is conserved
b.) [tex]V_{1}[/tex] = -1.075 m/s
c.) Same change of momentum.
COLLISION
We have different type of collision.
The parameters given in the question are:
- mass of the cart A [tex]M_{1}[/tex] = 200g = 0.2 kg
- initial velocity of cart A [tex]U_{1}[/tex] = 1.2 m/s
- final velocity of cart A [tex]V_{1}[/tex] = ?
- mass of the cart B [tex]M_{2}[/tex] = 350 g = 0.35 kg
- initial velocity of cart B [tex]U_{2}[/tex] = 2m/s
- final velocity of cart B [tex]V_{2}[/tex] = 0.7 m/s
a.) The reason this is considered as an isolated system of colliding bodies is because there is no loss of energy and momentum.
b.) To calculate the velocity of the cart A, let us use the formula below.
[tex]M_{1}[/tex][tex]U_{1}[/tex] - [tex]M_{2}[/tex][tex]U_{2}[/tex] = [tex]M_{1}[/tex][tex]V_{1}[/tex] - [tex]M_{2}[/tex][tex]V_{2}[/tex] ( since they are opposite to each other)
substitute all the parameters into the equation
0.2 x 1.2 - 0.35 x 2 = 0.2 x [tex]V_{1}[/tex] - 0.35 x 0.7
0.24 - 0.7 = 0.2[tex]V_{1}[/tex] - 0.245
-0.46 = 0.2[tex]V_{1}[/tex] - 0.245
0.2[tex]V_{1}[/tex] = -0.46 + 0.245
0.2[tex]V_{1}[/tex] = -0.215
[tex]V_{1}[/tex] = -0.215/0.2
[tex]V_{1}[/tex] = - 1.075 m/s
c.) Change in momentum of cart A = - 0.2 x 1.075 - 0.2 x 1.2
= -0.215 - 0.24
= -0.455 Kgm/s
Change in momentum of cart B = -0.245 + 0.7
= 0.455 Kgm/s
Therefore, they both have the same change of momentum but in opposite direction.
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