Physics Problems & Solutions: Linear Momentum

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Classical Mechanics - Linear Momentum

As shown in the picture, a ball of mass m suspended on the end of a wire, is released from height h and collides elastically, when it is at its lowest point, with a block of mass 2m at rest on a frictionless surface. After the collision, the ball rises to a final height equal to

A. 1/9 h
B. 1/8 h
C. 1/3 h
D. 1/2 h
E. 2/3 h

(GR9677 #07)

Solution:

Conservation of momentum of the system:

m_av_a+ m_bv_b = m_av_a' + m_bv_b'

Given:
m_a= m
m_b= 2m
v_b = 0

mv_a+ 0 = mv_a' + 2mv_b'
v_a = v_a' + 2v_b' (Eq.1)

Conservation of kinetic energy of the system:

½ m_av_a² + ½ m_bv_b² = ½ m_av_a'² + ½ m_bv_b'²
mv_a² + 0 = mv_a'² + 2mv_b'²
v_a² = v_a'² + 2v_b'² (Eq.2)

(Eq.1) → (Eq.2)
(v_a' + 2v_b')² = v_a'² + 2v_b'²
v_a'² + 4v_a'v_b' + 4v_b'² = v_a'² + 2v_b'²
4v_a'v_b' = 2v_b'² − 4v_b'²
2v_a'= − v_b'
v_b' = −2v_a' (Eq.3)

(Eq.3) → (Eq.1)
v_a = v_a' + 2v_b'
v_a = v_a' + 2(−2v_a')
v_a = − 3v_a' (Eq. 4)

For the pendulum, conservation of energy:

at the moment of the collision
U = T → m_agh = ½ m_av_a² → v_a = (2gh)^½

after the collision
U'= T' → m_agh' = ½ m_av_a'² → v_a' = (2gh')^½

Thus, (Eq. 4):
v_a = − 3v_a'
(2gh)^½ = − 3(2gh')^½
[(2gh)^½]² = [− 3(2gh')^½]²

h = 9h'
h' = ¹⁄₉h

Answer: A

Classical Mechanics - Linear Momentum

A helium atom, mass 4u travels with non relativistic speed v normal to the surface of a certain material, makes an elastic collision with an (essentially free) surface atom, and leaves in the opposite direction with speed 0.6v. The atom on the surface must be an atom of

A. Hydrogen, mass 1u
B. Helium, mass 4u
C. Carbon, mass 12u
D. Oxygen, mass 16u
E. Silicon, mass 28u

(GR9677 #20)

Solution:

m_a = 4u
v_a = v
v_b= 0
v_a' = − 0.6v

Conservation of momentum of the system:

m_av_a+ m_bv_b = m_av_a' + m_bv_b'
4uv = 4u(− 0.6v) + m_bv_b'
4uv = − 2.4uv + m_bv_b'
m_bv_b' = 6.4uv (Eq.1)

Conservation of kinetic energy of the system:

½ m_av_a² + ½ m_bv_b² = ½ m_av_a'² + ½ m_bv_b'²
4uv² = 4u(− 0.6v)² + m_bv_b'²
4uv² = 4u(0.36v²) + m_bv_b'²
m_bv_b'² = 4uv² − 1.44uv²
m_bv_b'² = 2.56uv² (Eq.2)

(Eq.1) → (Eq.2)
6.4 (v_b') = 2.56v
v_b' = (2.56/6.4)v = 0.4v (Eq.3)

(Eq.3) → (Eq.1)
m_b= 6.4uv / 0.4v = 16u

Answer: D

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

Classical Mechanics - Pendulum

Two small spheres of putty, A and B of mass M and 3M, respectively, hang from the ceiling on strings of equal length l. Sphere A is drawn aside so that it is raised to a height h₀ as shown above and then released. Sphere A collides with sphere B; they stick together and swing to a maximum height h equal to

A. (1/16) h₀
B. (1/8) h₀
C. (1/4) h₀
D. (1/3) h₀
E. (1/2) h₀

(GR8677 #5)

Solution:

Conservation of energy of A before and when it hits B:
$\begin{aligned} Mgh_0&=\frac{1}{2} Mv_A^2\\v_A&=\sqrt{2gh_0}\end{aligned}$

Conservation of momentum when and after collision:
$\begin{aligned} m_A v_A+m_B v_B&=(m_A+m_B)v_{AB}\\M\sqrt{2gh_0}+0&=4Mv_{AB}\\v_{AB}&=\frac{\sqrt{2gh_0}}{4}\end{aligned}$

Conservation of energy of A and B at h = 0 and h_max:
$\begin{aligned} \frac{1}{2} m_{AB} v_{AB}^2&=m_{AB} gh_{\max}\\\frac{1}{2} 4M\left (\frac{\sqrt{2gh_0}}{4} \right )^2&=4Mgh_{\max}\\h_{\max}&=\frac{1}{16} h_0\end{aligned}$

Answer: A

Classical Mechanics - Elastic Collision

A uniform stick of length L and mass M lies on a frictionless horizontal surface. A point particle of mass m approaches the stick with speed v on a straight line perpendicular to the stick that intersects the stick at one end, as shown above. After the collision, which is elastic, the particle is at rest. The speed V of the center of mass of the stick after the collision is

A. m/Mv
B. m/(M + m)v
C. √(m/M)v
D. √[m/(M + m)]v
E. 3m/Mv

(GR8677 #44)

Solution:

Conservation of momentum (elastic collision):
m_Av_A + m_B v_B = m_Av_A' + m_B v_B'
mv + 0 = 0 + MV
V = m/Mv

Answer: A

Quantum Mechanics - Eigenfunction

Which of the following is an eigenfunction of the linear momentum operator −iħ ∂/∂x with a positive eigenvalue ħk; i.e., an eigenfunction that describes a particle that is moving in free space in the direction of positive x with a precise value of linear momentum?

A. cos kx
B. sin kx
C. e^−ikx
D. e^ikx
E. e^−kx

(GR8677 #57)

Solution:

Eigenequation:

$\begin{aligned} \hat{p}\psi&=a\psi \\-i\hbar\frac{\partial \psi}{dx}&=\hbar k\psi \\\frac{\partial \psi}{dx}&=ik\psi \end{aligned}$

(A) FALSE
cos kx → ∂(cos kx)/∂x = sin kx

(B) FALSE
sin kx → ∂(sin kx)/∂x = cos kx

(C) FALSE
e^−ikx → ∂(e^−ikx)/∂x = −ik e^−ikx

(D) TRUE
e^ikx → ∂(e^ikx)/∂x = ik e^ikx

(E) FALSE
e^−kx→ ∂(e^kx)/∂x = k e^kx

Answer: D

Electromagnetism - Particle's Trajectory

A particle with charge q and momentum p is moving in the horizontal plane under the action of a uniform vertical magnetic field of magnitude B. Measurements are made of the particle's trajectory to determine the “sagitta” s and half-chord length l, as shown in the figure above. Which of the following expressions gives the particle's momentum in terms of q, B, s, and l? (Assume s ≪ l).

A. qBs²/2l
B. qBs²/l
C. qBl/s
D. qBl²/2s
E. qBl²/8s

(GR9677 #89)

Solution:

$\\F=qvB=\frac{mv^2}{R} \\\\qB=\frac{mv}{R}=\frac{p}{R} \rightarrow p=qBR \\\\R^2=l^2+(R-s)^2=l^2+R^2-2Rs+s^2 \\s\ll l \\R^2=l^2+R^2-2Rs \rightarrow l^2=2Rs \rightarrow R=l^2/2s \\\\\rightarrow p=qBR=qBl^2/2s$

Answer: D

Classical Mechanics - Inelastic Collision

In a non relativistic, one-dimensional collision, a particle of mass 2m collides with a particle of mass m at rest. If the particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision?

A. 0
B. 1/4
C. 1/3
D. 1/2
E. 2/3

(GR0177 #04)

Solution:

Conservation of Momentum (inelastic collision):

m_av_a + m_bv_b = (m_a + m_b)v'

2mv_a + m · 0 = (2m + m)v'

2v_a = 3v' → v' = 2v_a / 3

KE_i − KE_f = ( ½ m_av_a² + ½ m_bv_b²) − ½ ( m_a + m_b) (v')²

= mv_a² − ½ ( 2m + m) (2v_a / 3)²

= mv_a² − ⅔ mv_a²

= ⅓ mv_a²

Answer: C

Classical Mechanics - Conservation of Momentum

A particle of mass m is moving along the x-axis with speed v when it collides with a particle of mass 2m initially at rest. After the collision, the first particle has come to rest, and the second particle has split into two equal-mass pieces that move at equal angles θ

0 with the x-axis, as shown in the figure. Which of the following statements correctly describes the speeds of the two pieces?

A. Each piece moves with speed v
B. One of the pieces moves with speed v, the other moves with speed less than v.
C. Each piece moves with speed v/2
D. One of the pieces moves with speed v/2, the other moves with speed greater than v/2.
E. Each piece moves with speed greater than v/2

(GR0177 #55)

Solution:

Conservation of momentum:

$\begin{aligned} \\m_A v_A+m_B v_B&=m_A v_A'+2m_B' v_B' \\m v+0&=0+2\left(\frac{1}{2}\cdot 2m\right) v' \cos \theta \\v&=2v' \cos \theta \\v'&=\frac{v}{2\cos \theta}=\frac{1}{\cos \theta} \left( \frac{v}{2}\right ) \end{aligned}$

Answer: E

Condensed Matter - Effective mass

Lattice forces affect the motion of electrons in a metallic crystal, so that the relationship between the energy E and wave number k is not the classical equation E = ħ²k²/2m, where m is the electron mass. Instead, it is possible to use an effective mass m* given by which of the following?

A. $m^*=\frac{1}{2}\hbar^2 k \left( \frac{dk}{dE} \right)$

B. $m^*= \frac{\hbar^2 k^2}{\left( \frac{dk}{dE} \right)}$

C. $m^*= \frac{1}{2} \hbar^2 m^2 \left( \frac{d^2k}{dE^2} \right)^{1/3}$

D. $m^*= \frac{\hbar^2}{\left( \frac{d^2E}{dk^2} \right)}$

E. $m^*=\frac{1}{2} \hbar^2 m^2 \left( \frac{d^2E}{dk^2} \right)$

(GR9277 #97)

Solution:

F = m*a

F = dp/dt = ħ dk/dt (momentum, p = ħk)

a = dv_g/dt

Group velocity, v_g= dω/dk

Energy of electron, E = hf = ħω
ω = E/ħ

$v_g=\frac{d\omega}{dk} = \frac{1}{\hbar} \left( \frac{dE}{dk} \right)$

$a=\frac{dv_g}{dt}= \frac{1}{\hbar} \frac{d}{dt}\left( \frac{dE}{dk} \right)\frac{dk}{dk}= \frac{1}{\hbar} \left( \frac{d^2E}{dk^2} \right)\frac{dk}{dt}$

F = m*a

$\hbar \frac{dk}{dt} =m^*\cdot \frac{1}{\hbar} \left( \frac{d^2E}{dk^2} \right)\frac{dk}{dt}$

$m^*= \frac{\hbar^2}{\left( \frac{d^2E}{dk^2} \right)}$

Thermal Physics - Kinetic Theory Analysis

Using kinetic theory analysis, show that pressure of ideal gas is proportional to the average translational kinetic energy and number density.

$P=\frac{2}{3} \times \frac {1}{2} m\langle v^2\rangle \times \frac {N}{V}$

Solution:

Force: $F=ma=m \frac{\Delta v}{\Delta t}=\frac{\Delta p} {\Delta t}$

Total Force: $F=n\frac{\Delta p} {\Delta t}$

n = Number of collision in time Δt
Number of particles that moves in one direction highly likely is half of total particle, n = ½ N

Density: N/V
Volume of container : V = v_x ∆tA

$n= \frac{1}{2} N \frac{V}{V}=\frac{1}{2} \frac {N}{V} v_x \Delta tA$

Δp = Momentum transferred to wall per elastic collision.
Ideal gas → elastic collision → particle moves in x-direction with velocity v_x and bounces back with velocity −v_x.

$\Delta p=m[v_x-(-v_x ) ]=2mv_x$

Force:

$F= \left(\frac{1}{2} \frac {N}{V} v_x \Delta tA \right )\left(\frac{2mv_x }{\Delta t}\right )$

Pressure:

$P=\frac{F}{A}= \frac{\left(\frac{1}{2} \frac {N}{V} v_x \Delta tA \right )\left(2mv_x\right )}{\Delta tA}=\frac {N}{V}mv^2_x$

Average velocity, equal probability:

$\langle v^2 \rangle=\langle v_x^2 \rangle+\langle v_y^2 \rangle+\langle v_z^2 \rangle$
$\langle v_x^2 \rangle=\langle v_y^2 \rangle=\langle v_z^2 \rangle=\frac{1}{3}\langle v^2 \rangle$

Pressure:

$P=\frac{1}{3}\frac {N}{V}m\langle v^2\rangle$

In terms of average kinetic energy and number density, pressure:

$P=\frac{2}{3} \times \frac {1}{2} m\langle v^2\rangle \times \frac {N}{V}$

Pressure of ideal gas is proportional to the average translational kinetic energy and number density

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