# Failed physics test

I have a bunch of questions that I got wrong on a test and I'd appreciate some help on why my answers are incorrect:

1. A projectile is launched with an initial velocity of magnitude V at an angle t to the horizontal. Find the projectile’s maximum vertical displacement. (Ignore effects due to the air).

* KE = PE for the problem, the PE at the top is mgh and KE = 0, and opposite for the bottom, solving for H with KE = 0.5*mv^2 and PE = mgh gives
gh = .5v^2, solving for h gives v^2/(2*g), the sin(t) can be neglected till the end since the vertical copmonent of the velocity is what counts.

3. The coefficient of static friction between a box and a ramp is 0.5. The ramp’s incline angle (with the horizontal is 30°. If the box is placed at rest on the ramp, the box will.

* Friction force = .5mgcos(30)
Gravitation force = mgsin(30)
simplifying to .25g and g*sqrt(3)/2, .25 < sqrt(3)/2 gravitation is larger and the box will accelerate.

An electric dipole consists of a pair of point charges each of magnitude 4.0 nC separated by a distance of 2.0 cm. What is the electric field strength at the point midway between the charges?

=Since the charges are of same magnitude in charge and same polarity (and same distance at the centerpoint) then the field will have the same value from either side. The pull to the left from one charge will be exactly countered by the pull from the charge on the right and by superposition, the field cancels to 0.

=the terminals of a 12 V battery with negligible internal resistance are connected by a wire containing four resistors. Three of the resistors (2 ohms, 3 ohms, and 6 ohms) form a parallel combination, which is in series with the fourth resistor (1 ohm). Find the energy dissipated as heat by the 2-ohm resistor in 4 s.

The equivalent resistance is 1+1/(1/2+1/3+1/6) or which simplifies to 2 ohms, so the current in the circuit is 6 amps. The power over the resistor is then P = I V = 12*6*4s = 288. The multiple choice answers ranged from 8 to 96

An electron (mass = m, charge = – e) is projected with speed v upward, in the plane of page, into a regioncontaining a uniform magnetic field, B, that’s directed into the plane of the page. Describe the electron’s subsequent circular motion.

I guessed on this one trying to use some common sense, the options were arrangements of (mv) and (eB) as a ratio, and the direction of rotation. The rotation should be counter clockwise cuz of the right hand rule, and I figured as B got stronger the particle would deflect more (greatre force) so eB should inversly affect the radius. (Larger pull smaller radius). I guess not.

Traveling at an initial speed of 1.5 x 106 m/s, a proton enters a region of constant magnetic field of magnitude 1.5 T. If the proton’s initial velocity vector makes an angle of 30o with the magnetic field, compute the proton’s speed 4 s after entering the magnetic field.

= I had no idea on this one.

Thanks for the help.

ehild
Homework Helper
whozum said:
I have a bunch of questions that I got wrong on a test and I'd appreciate some help on why my answers are incorrect:

1. A projectile is launched with an initial velocity of magnitude V at an angle t to the horizontal. Find the projectile’s maximum vertical displacement. (Ignore effects due to the air).

* KE = PE for the problem, the PE at the top is mgh and KE = 0,

Wrong. The projectile keeps its horizontal velocity, so KE is not zero at the top.

3. The coefficient of static friction between a box and a ramp is 0.5. The ramp’s incline angle (with the horizontal is 30°. If the box is placed at rest on the ramp, the box will.

* Friction force = .5mgcos(30)
Gravitation force = mgsin(30)
simplifying to .25g and g*sqrt(3)/2, .25 < sqrt(3)/2 gravitation is larger and the box will accelerate.

The static friction is less than or equal to mu*Normal force. This maximum value is 0.5 mgcos(30)=sqrt3/4 gm=0.433 g*m. The component of gravitation force along the slope is mgsin(30)=0.5 g*m. The box accelerates.

An electric dipole consists of a pair of point charges each of magnitude 4.0 nC separated by a distance of 2.0 cm. What is the electric field strength at the point midway between the charges?

=Since the charges are of same magnitude in charge and same polarity

Wrong. The charges of an electric dipole are of opposite polarity.

=the terminals of a 12 V battery with negligible internal resistance are connected by a wire containing four resistors. Three of the resistors (2 ohms, 3 ohms, and 6 ohms) form a parallel combination, which is in series with the fourth resistor (1 ohm). Find the energy dissipated as heat by the 2-ohm resistor in 4 s.

The equivalent resistance is 1+1/(1/2+1/3+1/6) or which simplifies to 2 ohms, so the current in the circuit is 6 amps. The power over the resistor is then P = I V = 12*6*4s = 288.

The dissipated energy is I*V*t, that is OK, but the question was not the total energy: only the energy dissipated by the 2 ohm resistor was asked.
You have to calculate the voltage across this resistor. It is 6 V, as the other 6 V drops on the 1 ohm resistor in series. The power is U^2/R= 6^2/2=18 W, and the dissipated energy is 18*4 = 72 joule.
Do not forget the units when you give data !!!!

An electron (mass = m, charge = – e) is projected with speed v upward, in the plane of page, into a regioncontaining a uniform magnetic field, B, that’s directed into the plane of the page. Describe the electron’s subsequent circular motion.

I guessed on this one trying to use some common sense, the options were arrangements of (mv) and (eB) as a ratio, and the direction of rotation. The rotation should be counter clockwise cuz of the right hand rule, and I figured as B got stronger the particle would deflect more (greatre force) so eB should inversly affect the radius. (Larger pull smaller radius). I guess not.

The Lorentz force is e[vxB] charge times the vector product of the velocity and the magnetic field. This force presents the centripetal force that makes the charged particle move along a circle or radius R:
mv^2/R = evB --> R=(mv)/(eB). You are right about the effect of B.

Traveling at an initial speed of 1.5 x 106 m/s, a proton enters a region of constant magnetic field of magnitude 1.5 T. If the proton’s initial velocity vector makes an angle of 30o with the magnetic field, compute the proton’s speed 4 s after entering the magnetic field.

= I had no idea on this one.

It is the Lorentz force again. As this force is normal to the velocity, it does not effect the magnitude. So the speed stays the same, only the direction of the velocity changes.

ehild

ehild said:
Wrong. The projectile keeps its horizontal velocity, so KE is not zero at the top.
How would I solve this problem then?
The static friction is less than or equal to mu*Normal force. This maximum value is 0.5 mgcos(30)=sqrt3/4 gm=0.433 g*m. The component of gravitation force along the slope is mgsin(30)=0.5 g*m. The box accelerates.
Got it
Wrong. The charges of an electric dipole are of opposite polarity.

In the problem it shows the charge as 4 nC without sign, is it an inherent property of dipoles that the charges are opposite or am i right in assuming so from the problem?

The dissipated energy is I*V*t, that is OK, but the question was not the total energy: only the energy dissipated by the 2 ohm resistor was asked.
You have to calculate the voltage across this resistor. It is 6 V, as the other 6 V drops on the 1 ohm resistor in series. The power is U^2/R= 6^2/2=18 W, and the dissipated energy is 18*4 = 72 joule.
Do not forget the units when you give data !!!!

How do I calculate the voltage over that resistor? I know the resistance is 2 and voltage is 12.. 6 amps. I dont see where you get 6V. The equiv resistance of the parallel portion is 1ohm, and 1ohm for the separate resistor, how can I calculate both sets of resistors' voltage drop? Ohm's law doesnt seem to be helping me here.
The Lorentz force is e[vxB] charge times the vector product of the velocity and the magnetic field. This force presents the centripetal force that makes the charged particle move along a circle or radius R:
mv^2/R = evB --> R=(mv)/(eB). You are right about the effect of B.

If the answer is mv/eB why did I get it wrong then?

It is the Lorentz force again. As this force is normal to the velocity, it does not effect the magnitude. So the speed stays the same, only the direction of the velocity changes.

It says the proton enters the magnetic field at 30degrees, There is a copmonent of hte field that is parallel to the proton's velocity.

ehild
Homework Helper
whozum said:
How would I solve this problem then?

Separate the vertical and horizontal motions of the projectile.

In the problem it shows the charge as 4 nC without sign, is it an inherent property of dipoles that the charges are opposite or am i right in assuming so from the problem?

A dipole means opposite charges of equal magnitude at a distance of each other.

How do I calculate the voltage over that resistor?

Use Ohm's law.

I know the resistance is 2 and voltage is 12.. 6 amps. I dont see where you get 6V. The equiv resistance of the parallel portion is 1ohm, and 1ohm for the separate resistor, how can I calculate both sets of resistors' voltage drop? Ohm's law doesnt seem to be helping me here.

Why not? Those three resistors connected parallel can be replaced by one resistor of 1 ohm. This resistor is connected in series with the original 1 ohm resistor. You get 6 A current flowing out of the battery and flowing through both resistors. According to Ohm's law, the voltage drop on both 1 ohm resistor is 6 V.
If you connect resistors in parallel, the voltage is the same across each of them. So the voltage across the 2 ohm resistor is 6 also V.

If the answer is mv/eB why did I get it wrong then?

Have you said anything about v?
Well, and it is an electron, with -e charge. It will turn clockwise.

It says the proton enters the magnetic field at 30degrees, There is a copmonent of hte field that is parallel to the proton's velocity.

The parallel component of the magnetic field has no effect.

ehild

Last edited:
Romperstomper
whozum said:
In the problem it shows the charge as 4 nC without sign, is it an inherent property of dipoles that the charges are opposite or am i right in assuming so from the problem?

Electric fields between opposite charges will add, not subtract each other.

whozum said:
How do I calculate the voltage over that resistor? I know the resistance is 2 and voltage is 12.. 6 amps. I dont see where you get 6V. The equiv resistance of the parallel portion is 1ohm, and 1ohm for the separate resistor, how can I calculate both sets of resistors' voltage drop? Ohm's law doesnt seem to be helping me here.

6 amps go thorugh the total circuit. So 6 amps goes through the 1 ohm resistor and a total of 6 amps through all the resistors in parallel. The voltage through the parallel resistors is V = 6amps*1ohm. The voltage in resistors that are parallel are equal, so the voltage through the 2ohm resistor is 6 volts.