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2 years ago

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Kara Less was applying her makeup when she drove into South's busy parking lot last Friday morning. Unaware that Lisa Ford was s

topped in her lane 30 feet ahead, Kara rear-ended Lisa's rented Taurus. Kara's 1300-kg car was moving at 11 m/s and stopped in 0.14 seconds.a. Determine the momentum change of Kara's car.b. Determine the impulse experienced by Kara's car.c. Determine the magnitude of the force experienced by Kara's car.

2 answers:

exis [7]2 years ago

8 0

Answer

given,

Mass of Kara's car = 1300 Kg

moving with speed = 11 m/s

time taken to stop = 0.14 s

final velocity = 0 m/s

distance between Lisa ford and Kara's car = 30 m

a) change in momentum of Kara's car

Δ P = m Δ v

$\Delta P = m (v_f-v_i)$

$\Delta P = 1300 (0 - 11)$

Δ P = - 1.43 x 10⁴ kg.m/s

b) impulse is equal to change in momentum of the car

I = - 1.43 x 10⁴ kg.m/s

c) magnitude of force experienced by Kara

I = F x t

I is impulse acting on the car

t is time

- 1.43 x 10⁴= F x 0.14

F = -1.021 x 10⁵ N

negative sign represents the direction of force

Sever21 [200]2 years ago

7 0

Answer:

Explanation:

Given that,

Mass of kara car m =1300kg

Velocity at which kara car was moving Vi =11m/s

The car stopped after t= 0.14sec

Therefore the final velocity is Vf = 0m/s

a. What is the change in momentum?

Change is momentum can be determine using

∆p = ∆MV

∆p = M∆V

∆p = m(Vf-Vi)

∆p = 1300(0-11)

∆p = 1300×-11

∆p = —14,300 kgm/s

The change in momentum of kara's car is —14,300kgm/s.

b. Impulse felt be kara car?

Impulse can be determine by using Newton second law of motion

Ft = ∆p

Impulse is Ft

I = ∆p = -14,300kgm/s.

Then, the impulse felt by Kara's car is -14,300kgm/s

c. Magnitude of force experienced by Kara's car?

From the impulse formula,

Ft = ∆p

Therefore,

F = ∆p/t

F = -14,300/0.14

F = -102,142.9N.

F ≈ —102,143N

The force experienced by kara's is -102,143N.

We can also determined the deceleration of Kara's car

From Newton second law

F=ma

Then, a = F/m

a = -102,143/1300

a = -78.57m/s²

The negative sign show deceleration

We can also calculate the distance the car moved before coming to halt.

Using equation of motion

X =ut+½at²

X = 11×0.14 + ½(-78.57) × 0.14²

X = 1.54 —0.77

X = 0.77m

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