Question

a 1000 kilogram car strikes a tree at 30 km/hr and comes to a stop in 0.15s. Find the initial momentum and average force on the car while it is being stopped?

Answer 1

Answer on Question 81843, Physics, Mechanics, Relativity

Question:

A $1000 \, kg$ car strikes a tree at $30 \, km/h$ and comes to a stop in $0.15 \, s$. Find the initial momentum and average force on the car while it is being stopped?

Solution:

Let's first convert $km/h$ to $m/s$:

$$v_i = 30 \, \frac{km}{h} \cdot \frac{1000 \, m}{1 \, km} \cdot \frac{1 \, h}{3600 \, s} = 8.33 \, \frac{m}{s}.$$

We can find the initial momentum from the formula:

$$p_i = m v_i = 1000 \, kg \cdot 8.33 \, \frac{m}{s} = 8330 \, kg \cdot \frac{m}{s}.$$

Finally, we can find the average force on the car while it is being stopped from the definition of the impulse:

$$J = m \Delta v = F_{avg} \Delta t,$$ $$m(v_f - v_i) = F_{avg} \Delta t,$$ $$F_{avg} = \frac{m(v_f - v_i)}{\Delta t} = \frac{1000 \, kg \cdot \left(0 \, \frac{m}{s} - 8.33 \, \frac{m}{s}\right)}{0.15 \, s} = -55533 \, N.$$

The sign minus indicates that the average force on the car is directed opposite to the motion of the car.

Answer:

$$p_i = 8330 \, kg \cdot \frac{m}{s}.$$

$F_{avg} = 55533 \, N$, in opposite direction to the motion of the car.

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