Question

Checkpoint 4 1. Refer to Figure 1.25. Cart B, of mass 350 g, moves on the frictionless linear air track at 2 m. s-' to the left. Cart B strikes cart A, of mass 200 g, travelling in the opposite direction at 1,2 m. st. After the collision, cart B continues in its original direction at 0,7 m. s', a)Why is this considered an isolated system of colliding bodies? b) Calculate the velocity of cart A after the collision. c) How does the change in momentum of each cart compare? Check your answer using calculations. Cart B Cart A 2 m. s1 1.2 m. s 1 S" 200 g 350 g​

Answer 1

From the knowledge of linear momentum,

a.) momentum is conserved

b.) $V_{1}$ = -1.075 m/s

c.) Same change of momentum.

COLLISION

We have different type of collision.

• Elastic

• inelastic

• perfectly elastic

• perfectly inelastic

The parameters given in the question are:

• mass of the cart A $M_{1}$ = 200g = 0.2 kg

• initial velocity of cart A $U_{1}$ = 1.2 m/s

• final velocity of cart A $V_{1}$ = ?

• mass of the cart B $M_{2}$ = 350 g = 0.35 kg

• initial velocity of cart B $U_{2}$ = 2m/s

• final velocity of cart B $V_{2}$ = 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.

$M_{1}$$U_{1}$ - $M_{2}$$U_{2}$ = $M_{1}$$V_{1}$ - $M_{2}$$V_{2}$ ( 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 $V_{1}$ - 0.35 x 0.7

0.24 - 0.7 = 0.2$V_{1}$ - 0.245

-0.46 = 0.2$V_{1}$ - 0.245

0.2$V_{1}$ = -0.46 + 0.245

0.2$V_{1}$ = -0.215

$V_{1}$ = -0.215/0.2

$V_{1}$ = - 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.

Learn more about Momentum here: brainly.com/question/402617