Physics Problems & Solutions: #65

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Classical Mechanics - Conservation of Momentum

A man of mass m on an initially stationary boat gets off the boat by leaping to the left in an exactly horizontal direction. Immediately after the leap, the boat of mass M is observed to be moving to the right at speed v. How much work did the man do during the leap (both on his own body and on the boat)?

A. ½ Mv²
B. ½ mv²
C. ½ (M + m)v²
D. ½ (M + M²/m)v²
E. ½ [Mm/(M + m)]v²

(GR9677 #65)

Solution:

Conservation of momentum:

$mv_0=Mv \rightarrow v_0=\frac{M}{m}v$

The man does work on both himself and the boat:

$\begin{aligned} \\W&=KE_{\text{total}} \\&=\frac{1}{2}mv_0^2+\frac{1}{2}Mv^2 \\&=\frac{1}{2}m\left ( \frac{M}{m}v \right )^2+\frac{1}{2}Mv^2 \\&=\frac{1}{2}\left ( \frac{M^2}{m}+M \right )v^2 \end{aligned}$

Answer: D

Electromagnetism - Coulomb's Law

Two point charges with the same charge +Q are fixed along the x-axis and are a distance 2R apart as shown. A small particle with mass m and charge –q is placed at the midpoint between them. What is the angular frequency ω of small oscillations of this particle along the y-direction?

A. $\frac{Qq}{2\pi \epsilon_\theta mR^2}$

B. $\frac{Qq}{4\pi \epsilon_\theta mR^2}$

C. $\frac{Qq}{2\pi \epsilon_\theta mR^3}$

D. $\left (\frac{Qq}{4\pi \epsilon_\theta mR^2} \right )^{\frac{1}{2}}$

E. $\left (\frac{Qq}{2\pi \epsilon_\theta mR^3} \right )^{\frac{1}{2}}$

(GR9277 #65)

Solution:

Check the unit, ω → sec⁻¹

with
Q = q → Coulomb
R = meter
m = kg
ϵ₀ = coulomb² / (newton · meter²) = coulomb²·sec²/ (kg · meter³)

A. coulomb² · kg · meter³/ (coulomb² · sec² · kg · meter²) = meter/sec²
→ FALSE.

B. Identical unit with A. → FALSE

C. 1/sec² → FALSE

D. √m/sec → FALSE

E. 1/sec → TRUE

Answer: E

Calculation

Coloumb's Law: $F=k\frac{Qq}{r^2}$

In this case,

$\vec{F}=\frac{1}{4\pi \epsilon_0}\frac{2Q(-q)}{R^2} \text{ sin} \theta \text{ }\hat j =-\frac{1}{2\pi \epsilon_0}\frac{Qq}{R^2} \text{ sin} \theta\text{ }\hat j$

(Electric fields from both Qs are in x-and y-direction, but in x-axis, they cancel each other.)

For small θ, sin θ ≈ tan θ = y/R

Hooke's Law in y-axis: $F=m\ddot{y}=-ky$

Angular frequency, $\omega = \sqrt{k/m} = - \ddot{y}/y\rightarrow \ddot{y} = -\omega^2 y$

$\begin{aligned} m\ddot{y}&=-\frac{1}{2\pi \epsilon_0}\frac{Qq}{R^2} \frac{y}{R} \\ -m\omega^2 y &=-\frac{1}{2\pi \epsilon_0}\frac{Qq}{R^2} \frac{y}{R} \\\omega^2 &= \frac{Qq}{2\pi \epsilon_0mR^3} \\\omega &= \left ( \frac{Qq}{2\pi \epsilon_0mR^3} \right )^{\frac{1}{2}} \end{aligned}$

Electromagnetism - Biot Savart's Law

A current i in a circular loop of radius b produces a magnetic field. At a fixed points far from the loop, the strength of the magnetic field is proportional to which of the following combinations of i and b?

A. ib
B. ib²
C. i²b
D. i/b
E. i/b²

(GR8677 #65)

Solution:

Applying the concept of magnetic dipole moment:
μ = a vector quantity associated with the torque exerted by an external magnetic field on a current carrying coil

μ = IA = Iπr² = iπb²

Answer: B

Note: click HERE for complete calculation to show that magnetic field at a fixed distance x far away from a circular loop: $B= \frac{\mu_0 IR^2}{2x^3}$

Thermal Physics - Heat Capacity

$C=3kN_A \left( \frac{hv}{kT} \right)^2 \frac{e^{hv/kT}}{(e^{hv/kT}-1)^2}$
Einstein’s formula for the molar heat capacity C of solids is given above. At high temperatures, C approaches which of the following?

A. 0
B. 3kN_A(hv/kT)
C. 3kN_Ahv
D. 3kN_A
E. N_Ahv

(GR0177 #65)

Solution:

At high temperature → Classical approach, there’s no hv involved.

Answer: D