Question 1

A car traveling at 11.6 meters per second crashes into a barrier and stops in 0.287 meters. What force must be exerted on a child of mass 21.2 kilograms to stop him or her in the same time as the car? Include units in your answer. Answer must be in 3 significant digits. Hint: This is an impulse-momentum theorem problem.

Question 2

$\begin{gathered} \text{For car} \\ v_1=11.6\text{ m/s} \\ \Delta x=0.287m \\ v_2=\text{ 0 m/s} \\ t=\text{?} \\ To\text{ find t} \\ v^2_2=v^2_1+2a\Delta x \\ \text{Solving a} \\ v^2_2-v^2_1=2a\Delta x \\ a=(v^2_2-v^2_1)/(2\Delta x) \\ a=((0m/s)^2-(11.6m/s)^2)/(2(0.287m)) \\ a=(0m^2/s^2-134.56m^2/s^2)/(0.574m) \\ a=(-134.56m^2/s^2)/(0.574m) \\ a=-234.42m/s^2 \\ \text{Then} \\ t=(v_2-v_1)/(a) \\ t=\frac{0\text{ m/s-11.6m/s}}{-234.42m/s^2} \\ t=\frac{\text{-11.6m/s}}{-234.42m/s^2} \\ t=0.0495\text{ s} \\ \text{For child} \\ m=21.2\text{ kg} \\ v_1=11.6\text{ m/s} \\ v_2=\text{ 0 m/s} \\ t=0.0495\text{ s} \\ F=\text{?} \\ F=(P_2-P_1)/(t) \\ P_2-P_1=mv_2-mv_1=m(v_2-v_1),\text{ then} \\ F=(m(v_2-v_1))/(t) \\ F=\frac{(21.2kg)(0\text{ m/s-11.6m/s})}{0.0495\text{ s}} \\ F=\frac{(21.2kg)(-11.6\text{ m/s})}{0.0495\text{ s}} \\ F=\frac{-245.92\operatorname{kg}\text{ m/s}}{0.0495\text{ s}} \\ F=-4968\text{ N} \\ \text{The force to stop the child is 4968 N. The negative means } \\ \text{the force is opposite to the movement} \end{gathered}$

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