Question

A 2000kg car hit a wall with 20ms-¹ initial velocity. After 0.15s hitting time, the car got 1.3ms-¹ velocity backward. The question is what is impulse of the force of the wall on that car?

Answer 1

A 2000kg car hit a wall with 20ms⁻¹ initial velocity. After 0.15s hitting time, the car got 1.3ms⁻¹ velocity backward. The question is what is impulse of the force of the wall on that car?

The second law states that the net force on an object is equal to the rate of change (that is, the derivative) of its linear momentum $p$ in an inertial reference frame:

$$\vec{F} = \frac{d\vec{p}}{dt} = \frac{\Delta\vec{p}}{\Delta t}$$

where $\Delta\vec{p}$ – change of momentum $\vec{p} = m\vec{v}$

$\Delta t$ – time of interaction.

In our case change of momentum equals:

$$\Delta\vec{p} = 2000\,kg \cdot \left(20\,\frac{m}{s} - \left(-1.3\,\frac{m}{s}\right)\right) = 2000\,kg \cdot 21.3\,\frac{m}{s} = 42600\,kg\,\frac{m}{s}$$

Therefore, force equals:

$$F = \frac{42600\,kg\,\frac{m}{s}}{0.15\,s} = 284000\,N$$

Answer: 284000 N