Question

Derive an expression relating impulse and linear momentum. In a safety test, a car of mass
1000 kg is driven into a brick wall. Its bumper behaves like a spring (k = 5 × 106 Nm−1) and is
compressed by a distance of 3 cm as the car comes to rest. Determine the initial speed of the car.

Answer 1

Derive an expression relating impulse and linear momentum. In a safety test, a car of mass $1000\,\mathrm{kg}$ is driven into a brick wall. Its bumper behaves like a spring ($k = 5 \times 106\,\mathrm{Nm}^{-1}$) and is compressed by a distance of $3\,\mathrm{cm}$ as the car comes to rest. Determine the initial speed of the car.

Impulse $J$ produced from time $t_1$ to $t_2$ is defined to be:

$$J = \int_{t_1}^{t_2} F\,dt$$

where $F$ is the force applied from $t_1$ to $t_2$.

From Newton's second law, force is related to momentum $p$ by:

$$F = \frac{dp}{dt}$$

Therefore:

$$J = \int_{t_1}^{t_2} F\,dt = \int_{t_1}^{t_2} \frac{dp}{dt}\,dt = \int_{t_1}^{t_2} dp = \Delta p$$

The energy conservation law:

$$\frac{mv^2}{2} = \frac{k\Delta l^2}{2}$$

m – mass of car

v – initial speed of car

$\Delta l$ – deformation of bumper

Therefore:

$$v = \sqrt{\frac{k\Delta l^2}{m}} = \sqrt{\frac{5 \cdot 10^6 \cdot \frac{N}{m} (0.03\,m)^2}{1000\,kg}} = 2.12\,\frac{\mathrm{m}}{\mathrm{s}}$$

Answer: the initial speed of the car equals $2.12\,\frac{\mathrm{m}}{\mathrm{s}}$