Solving Two Crashing Cars Problem - 11.3152 m/s

[KMjuniormint5]

You testify as an "expert witness" in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ = 12.0°, that the cars were separated by distance d = 25.5 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.0 m/s.

(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?

how I went about the problem:

a = acceleration only in x direction

for car A:
Fnet = mAa --> WA2 - Fk = (mA)a --> (mA)(g)(sin12.0) - \mu_k (mA)(g) = (mA)a

now i divided everything by (mB) to get final equation of:
(g)(sin12.0) - \mu_k (g) = a



now since constant acceleration I plugged a into the equation of

V^2 = 2a(delta x) + Vo^2 to get --> V^2 = 2((g)(sin12.0) - \mu_k (g))(x) =Vo^2

plugged in everything I know:

v^2 = 2 (2(9.8sin12.0 - ((0.6)(9.8)))(25.5) - 18^2

then take the square root to get v = 11.3152m/s

why is this not right?

 

[KMjuniormint5]

http://www.webassign.net/hrw/W0106-N.jpg <--- there is a picture if you need one

 

[stewartcs]

You testify as an "expert witness" in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ = 12.0°, that the cars were separated by distance d = 25.5 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.0 m/s.

(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?

how I went about the problem:

a = acceleration only in x direction

for car A:
Fnet = mAa --> WA2 - Fk = (mA)a --> (mA)(g)(sin12.0) - \mu_k (mA)(g) = (mA)a

now i divided everything by (mB) to get final equation of:
(g)(sin12.0) - \mu_k (g) = a



now since constant acceleration I plugged a into the equation of

V^2 = 2a(delta x) + Vo^2 to get --> V^2 = 2((g)(sin12.0) - \mu_k (g))(x) =Vo^2

plugged in everything I know:

v^2 = 2 (2(9.8sin12.0 - ((0.6)(9.8)))(25.5) - 18^2

then take the square root to get v = 11.3152m/s

why is this not right?

This belongs in the HOMEWORK section...see sticky at top of forum.

However, to get you started...Fnet = ma means the vector sum of the x AND y components, hence...

mg*sin(theta) - mu*mg*cos(theta) = ma

which gives...

a = g*sin(theta) - mu*g*cos(theta)

See if you can take it from here...

 

[KMjuniormint5]

but if on an incline. . .and we make the incline x-axis and the normal force the y-axis we do not have any acceleration in the y direction

 

[Doc Al]

how I went about the problem:

a = acceleration only in x direction

for car A:
Fnet = mAa --> WA2 - Fk = (mA)a --> (mA)(g)(sin12.0) - \mu_k (mA)(g) = (mA)a

You made a mistake in calculating the force of friction. (As stewartcs has already pointed out.) What's the normal force between car and incline? It's not mg!

 

[stewartcs]

but if on an incline. . .and we make the incline x-axis and the normal force the y-axis we do not have any acceleration in the y direction

There is acceleration in the y direction, gravity always points toward the center of the earth. So even if you are on an incline, and you make your coordinate system relative to the incline, gravity will still act toward the center of the earth. The force vector is pointing toward the center of the earth, not into the incline. The y component is pointing into the incline. This will be the normal force that will be used to find the opposing frictional force of the car.

Hence, the normal force is not the entire weight of the car, but rather the y component. In this case mg*cos(theta) instead of mg like you had in your original equation.