A 1.0-kg standard cart A collides with a 0.10-kg cart B. The x component of the velocity of cart A is +0.70 m/s before the collision and +0.50 m/s after the collision. Cart B is initially traveling toward cart A at 0.40 m/s , and after the collision the x component of its velocity is +1.6 m/s .

Part A: What is the x component of the change in the momentum of A?
Part B: What is the x component of the change in the momentum of B?
Part C: What is the sum of these two x components of the changes in momentum?

Question

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Respuesta :

abidemiokin abidemiokin · Answer 1 · 2020-04-23T08:09:28+02:00

Answer:

Part A

-0.2kgm/s

Part B

+0.12m/s

Part C

-0.08kgm/s

Explanation

Momentum of a body is defined as the product of mass of the body band its velocity.

Momentum = mass × velocity

Change in momentum = m(v-u)

v is the final velocity of the body

u is the initial velocity of the body.

For cart A:

Mass m = 1.0kg

Initial velocity u = +0.70m/s(velocity before collision)

Final velocity' v = +0.5m/s(velocity after collision)

∆Momentum = 1(0.5-0.7)

= 1(-0.2)

= -0.2kgm/s

The x component of the change in the momentum of A is -0.2kgm/s

For cart B:

Similarly;

∆momentum = 0.1(1.6-0.4)

∆momentum = 0.1×1.2

∆momentum = +0.12kgm/s

The x component of the change in the momentum of B is +0.12kgm/s

Part C:

The sum of these two x components of the changes in momentum = (∆momentum)A + (∆momentum)B

= -0.2+0.12

= -0.08kgm/s