Question

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

Answer 1

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

Answer 2

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