Question

You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light when it was hit from behind by car A, of mass 1300 kg. The cars locked bumpers during the collision and slid to a stop. Measurements of the skid marks left by the tires showed them to be 7.25 m long, and inspection of the tire tread revealed that the coefficient of kinetic friction between the tires and the road was 0.65.a) What was the speed of car A just before the collision? b) If the speed limit was 35mph, was car A speeding, and if so, by how many miles per hour was it exceeding the speed limit? c) Explain why we can classify the collision as completely inelastic d) Calculate the total kinetic energy before and after the collision

Answer 1

Answer:

a). va=17.23 $\frac{m}{s}$ or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

$m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65$

The motion is an inelastic collision so

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

The force of the motion is contrarest by the force of friction so

$F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}$

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

$x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s$

So the velocity final can be find using this time

$v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}$

a).

Replacing in the first equation the final velocity can find the initial velocity

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

$v_{b}=0$

$v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}$

b).

$35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}$

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

$Ek=\frac{1}{2}*m*(v)^{2}\\  Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ$

Answer 2

The speed of car A before the collision occured will be 48.30 miles per hour.

How to calculate the speed.

From the information given, the value of v through the work energy theorem will be:

= 2 × ✓0.65 × ✓9.8 × ✓7.15

= 9.54m/s

Then, the law of conservation of momentum will give the value of initial velocity to be:

= [(1900 + 1500)(9.54)] - [1900 × 0] / 1500

= 21.6

Therefore, the speed will be:

= 21.6 × 0.000621 × 3600

= 48.3 miles per hour

If the speed limit was 35mph, car A was speeding and the limit exceeded will be:

= 48.3 - 35

= 13.3

Since friction is involved, the energy would not be conserved. Therefore, the collision will be inelastic.

The total kinetic energy before the collision will be

= 1/2mv²A + 1/2mv²B

= 1/2(1500)(21.6)² + 1/2(1900)(0)

= 350kJ

The total kinetic energy after the collision will be

= 1/2(mA + mB)²

= 1/2(1500 + 1900)(9.54)

= 155kJ

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