Question

1. A car has a total mass of 1200 kg and is traveling at 100 km per hour when the driver experience is a brake failure and collides with the barrels. Calculate the change in momentum it will experience whilst coming to a standstill.2. According to GSU's HyperPhysics Project this crash would have been fatal for an average 80 kg person. The safety zone in terms of momentum, ranges from 0 to 1,000 kg per metre per second. Determine the minimum velocity the car can slow down to during a collision with the barrels without the crash being fatal.

Answer 1

Question 1.

Given:

Mass = 1200 kg

Velocity = 100 km per hour.

Let's find the change in momentum it will expperience.

To find the change in momentum, apply the formula:

$\Delta p=mv_f-mv_i$

Where:

Δp is the change in momentum

m is the mass of the car = 1200 kg

vf is the final velocity

vi is the initial velocity.

Here, the final velocity is = 0 m/s

The initial velocity is = 100 km/h

Let's convert the initial velocity to m/s.

Where:

1 m/s = 3.6 km/h

100 km/h = 100/3.6 = 27.78 m/s

Input the values into the formula and solve for Δp.

We have:

$\begin{gathered} \Delta p=(1200\times0)-(1200\times27.78) \\ \\ \Delta p=0-3333.33 \\ \\ \Delta p=-33333.33kg.m\text{ /s} \end{gathered}$

Therefore, the change in momentum is 33333.33 kg.m/s

Question 2.

Since the safety zone for momentum is 0 to 1000 kg.m/s, to find the minimum velocity of the car, substitute 1000 kg.m/s for Δp and solve for the final velocity vf

We have:

$\begin{gathered} \Delta p=m_{}v_f-mv_i \\ \\ 1000=(1200v_f)-(1200\times27.78) \\ \\ 1000=1200v_f-33333.33_{} \\ \\ v_f=\frac{1000-33333.33}{1200} \\ \\ v_f=-26.94\text{ m/s} \end{gathered}$

Therefore, the minimum velocity the car can slow down to a velocity is 26.94 m/s.

ANSWER:

(a). 33333.33 kg/m.s

(b). 26.94 m/s