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QuestionHomework 10 Begin Date: 11/3/2024 12:01:00 Aum Problem 3: ( 10%10 \% of Assignment Value) Suppose a child drives a bumper car head on into the side rail, which exerts a force of 3900 N on the car for 0.55 s . Use the initial direction of the car's motion as the positive direction. - Part (a) ✓\checkmark
What impulse, in kilogram meters per second, is imparted to the car by this force? J=−2145J=−2145\begin{array}{l} J=-2145 \\ J=-2145 \end{array} ✓\checkmark Correct!
Part (b) Find the horizontal components of the final velocity of the bumper car, in meters per second, if its initial velocity was 2.5 m/s2.5 \mathrm{~m} / \mathrm{s} and the car plus driver have a mass of 190 kg . You may neglect friction between the car and floor. vf=−13.79v_{\mathrm{f}}=-13.79 \begin{tabular}{lr} \hline Grade Summary & \\ \hline Deductions & 0%0 \% \\ Potential & 100%100 \% \end{tabular} Submit Hint Feedback I give up!
3 Submission(s) Remaining Hints: 0%‾\underline{0 \%} deduction per hint. Hints remaining: 2 Feedback: 2 for a 0%0 \% deduction Check your signs. What should the sign of initial momentum be? Submission History All Date times are displayed in Eatern Standard Time Red submission date times indicate late work. \begin{tabular}{|c|c|c|c|c|c|} \hline & Date & Time & Answer & Hints & Feedback \\ \hline 1 & Nov 16, 2024 & 1.26 PM & vp=13.79v_{p}=13.79 & & Check your signs. What should the sign of impulse imparted be? \\ \hline 2 & Nov 16, 2024 & 1:31 PM & vf=−13.79vf=−13.79\begin{array}{l} v_{\mathrm{f}}=-13.79 \\ v_{\mathrm{f}}=-13.79 \end{array} & & Check your signs. What should the sign of initial momentum be? \\ \hline \end{tabular} All content O 2024 Expert TA, LLC

Studdy Solution

STEP 1

1. The force exerted on the car is 3900 N.
2. The duration of the force is 0.55 seconds.
3. The initial velocity of the car is 2.5 m/s.
4. The mass of the car plus driver is 190 kg.
5. Friction between the car and floor is negligible.
6. The initial direction of the car's motion is considered positive.

STEP 2

1. Calculate the impulse imparted to the car.
2. Determine the change in momentum.
3. Calculate the final velocity of the bumper car.

STEP 3

Calculate the impulse imparted to the car using the formula for impulse:
J=F⋅t J = F \cdot t
where F=3900 N F = 3900 \, \text{N} and t=0.55 s t = 0.55 \, \text{s} .
J=3900 N×0.55 s=2145 Ns J = 3900 \, \text{N} \times 0.55 \, \text{s} = 2145 \, \text{Ns}
Since the force is exerted in the opposite direction of the initial motion, the impulse is negative:
J=−2145 Ns J = -2145 \, \text{Ns}

STEP 4

Determine the change in momentum. The impulse is equal to the change in momentum:
Δp=J \Delta p = J
Δp=−2145 Ns \Delta p = -2145 \, \text{Ns}

STEP 5

Calculate the final velocity of the bumper car using the relationship between impulse, change in momentum, and velocity:
Δp=m⋅(vf−vi) \Delta p = m \cdot (v_f - v_i)
where m=190 kg m = 190 \, \text{kg} , vi=2.5 m/s v_i = 2.5 \, \text{m/s} , and vf v_f is the final velocity.
Substitute the known values:
−2145=190⋅(vf−2.5) -2145 = 190 \cdot (v_f - 2.5)
Solve for vf v_f :
−2145=190vf−475 -2145 = 190v_f - 475
Add 475 to both sides:
−2145+475=190vf -2145 + 475 = 190v_f
−1670=190vf -1670 = 190v_f
Divide both sides by 190:
vf=−1670190 v_f = \frac{-1670}{190}
vf=−8.79 m/s v_f = -8.79 \, \text{m/s}
The final velocity of the bumper car is:
−8.79 m/s \boxed{-8.79 \, \text{m/s}}

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