# Change in momentum?

## A car with mass 1500kg travelling at 80km/h comes to complete stop in 4 seconds. What is the resulting change in momentum? Now assume the same car was in an accident and get from 80kmh to a complete stop in .25 seconds. Change in momentum? I know the formula for momentum is p=mv but I'm not sure how to calculate this.

Oct 30, 2017

In both cases, $\Delta p = - 3000 k g \cdot \frac{m}{s}$

#### Explanation:

Momentum is typically in units of kg*m/s. Therefore the 80 km/h should be converted to m/s before going on to momentum.
$80 \frac{k m}{h} \cdot \left(\frac{1000 m}{1 k m}\right) \cdot \left(\frac{1 h}{3600 s}\right) = 2.22 \frac{m}{s}$

Yes, p=mv. The initial momentum is
${p}_{i} = 1500 k g \cdot 2.222 \frac{m}{s} = 3333 k g \cdot \frac{m}{s}$

Since the final velocity is zero,
${p}_{f} = 0$

The change is
$\Delta p = {p}_{f} - {p}_{f} = 0 - 3333 k g \cdot \frac{m}{s} = - 3333 k g \cdot \frac{m}{s}$
Because of the rules regarding the number of significant digits, the answer to this part should be given as -3000 kg*m/s.

In the case of the accident , the initial momentum is
${p}_{i} = 1500 k g \cdot 2.222 \frac{m}{s} = 3333 k g \cdot \frac{m}{s}$

Since the final velocity is zero,
${p}_{f} = 0$

The change is ${p}_{f} - {p}_{f} = 0 - 3333 k g \cdot \frac{m}{s} = - 3333 k g \cdot \frac{m}{s}$
Because of the rules regarding the number of significant digits, the answer to this part should be given as -3000 kg*m/s.

The times that were given in the question were not relevant to what the question asked for. However, those times would be used if it also asked for what the average force was during those 2 cases.

Additional analysis: The average force during those 2 cases:

Change in momentum calculated above is called impulse. Another formula for impulse is calculated by multiplying average force by the time of application of that force. Both methods yield impulse, so we can write
$\Delta p = {F}_{\text{ave}} \cdot t$

We know that $\Delta p = - 3000 k g \cdot \frac{m}{s}$, so we can solve for ${F}_{\text{ave}}$.

The case of breaking to a stop in 4 s:

$\Delta p = - 3000 k g \cdot \frac{m}{s} = {F}_{\text{ave}} \cdot 4 s$

${F}_{\text{ave}} = \frac{- 3000 k g \cdot \frac{m}{s}}{4 s} = - 750 k g \cdot \frac{m}{s} ^ 2 = - 750 N$

The case of an accident stopping the car in 0.25 s:

$\Delta p = - 3000 k g \cdot \frac{m}{s} = {F}_{\text{ave}} \cdot 0.25 s$

${F}_{\text{ave}} = \frac{- 3000 k g \cdot \frac{m}{s}}{0.25 s} = - 12 , 000 k g \cdot \frac{m}{s} ^ 2 = - 12 , 000 N$

I hope this helps,
Steve