Question

A car of mass 120 kg is moving at 100 m/s. If it slows down to 40 m/s in 50 seconds:

A. The impulse and force that are being provided by the breaking mechanism is when the mechanism is brought to a stop and then “throw it back.” Impulses are greater when bouncing occurs.

Answer 1

Answer on Question #44596 – Physics – Mechanics | Kinematics | Dynamics

A car of mass $120\,\mathrm{kg}$ is moving at $100\,\mathrm{m/s}$. If it slows down to $40\,\mathrm{m/s}$ in 50 seconds:

A. The impulse and force that are being provided by the breaking mechanism is when the mechanism is brought to a stop and then "throw it back." Impulses are greater when bouncing occurs.

Solution:

$\mathrm{m} = 120\,\mathrm{kg}$ – mass of the car;

$\mathrm{v}_1 = 100\,\frac{\mathrm{m}}{\mathrm{s}}$ – initial velocity of the car;

$\mathrm{v}_2 = 40\,\frac{\mathrm{m}}{\mathrm{s}}$ – final velocity of the car;

$t = 50\,\mathrm{s}$ – deceleration time;

The impulse of force can be extracted and found to be equal to the change in momentum of an object provided the mass is constant:

$$\begin{array}{l} \text{Impulse} = \mathrm{m}\,\Delta\mathrm{v} = \mathrm{m}\,\mathrm{v}_2 - \mathrm{m}\,\mathrm{v}_1 = \mathrm{m}(\mathrm{v}_2 - \mathrm{v}_1) = 120\,\mathrm{kg} \cdot \left(40\,\frac{\mathrm{m}}{\mathrm{s}} - 100\,\frac{\mathrm{m}}{\mathrm{s}}\right) \\ = -7200\,\mathrm{N} \cdot \mathrm{s} \end{array}$$

Formula for the impulse:

$$\mathrm{F} = \frac{\text{Impulse}}{\mathrm{t}} = \frac{-7200\,\mathrm{N} \cdot \mathrm{s}}{50\,\mathrm{s}} = -144\,\mathrm{N}$$

We have a minus sign before force and impulse because the direction of force and impulse is opposite to the direction of motion (deceleration) of the car.

Answer: Impulse = $-7200\,\mathrm{N} \cdot \mathrm{s}$; F = $-144\,\mathrm{N}$.

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