Question

practice it
use the worked example above to help you solve this problem. in a crash test, a car of mass ( 1.49\times10^{3}\text{ kg} ) collides with a wall and rebounds as shown in the figure. the initial and final velocities of the car are ( v_{i}=-15.2\text{ m/s} ) and ( v_{f}=2.08\text{ m/s} ), respectively. if the collision lasts for ( 0.158\text{ s} ), find the following.
(a) the impulse delivered to the car due to the collision
( \text{kg}cdot\text{m/s} )
(b) the size of and direction of the average force exerted on the car (indicate the direction with the sign of your answer.)
( \text{n} )
exercise
hints: getting started | im stucki
use the values from practice it to help you work this exercise. suppose the car doesnt rebound off the wall, but the time interval of the collision remains at ( 0.158\text{ s} ). in this case, the final velocity of the car is zero. find the average force exerted on the car. (indicate the direction with the sign of your answer.)
( \text{n} )

Explanation:

Step1: Calculate the impulse (for part a)

Impulse \(J\) is given by the change in momentum \(\Delta p = m(v_f - v_i)\).
Given \(m = 1.49\times10^{3}\text{ kg}\), \(v_i=- 15.2\text{ m/s}\), \(v_f = 2.08\text{ m/s}\)

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Step2: Calculate the average force (for part b)

We know that \(J = F_{avg}\Delta t\), so \(F_{avg}=\frac{J}{\Delta t}\)
Given \(\Delta t=0.158\text{ s}\) and \(J = 2.57\times10^{4}\text{ kg}\cdot\text{m/s}\)

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Step3: Calculate the average force for the exercise

For the exercise, \(m = 1.49\times10^{3}\text{ kg}\), \(v_i=-15.2\text{ m/s}\), \(v_f = 0\text{ m/s}\), \(\Delta t = 0.158\text{ s}\)
First, find the impulse \(J=m(v_f - v_i)=1.49\times10^{3}(0-(-15.2))=1.49\times10^{3}\times15.2 = 2.26\times10^{4}\text{ kg}\cdot\text{m/s}\)
Then, using \(F_{avg}=\frac{J}{\Delta t}\)

$$
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Answer:

(a) \(2.57\times 10^{4}\text{ kg}\cdot\text{m/s}\)
(b) \(1.63\times 10^{5}\text{ N}\)
Exercise: \(1.43\times 10^{5}\text{ N}\)