Physics Problems & Solutions: Conservation Laws

Showing posts with label Conservation Laws. Show all posts

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

Nuclear & Particle Physics - Beta Decay

When the beta decay of ⁶⁰Co nuclei is observed at low temperature in a magnetic field that aligns spins of the nuclei, it is found that the electrons are emitted preferentially in a direction opposite to the ⁶⁰Co spin direction. Which of the following invariances is violated by this decay

A. Gauge invariance
B. Time invariance
C. Translation invariance
D. Reflection invariance
E. Rotation invariance

(GR9677 #34)

Solution:

Wu experiment, 1957: beta decay of Cobalt-60 shows that electrons are emitted preferentially in a direction opposite to the ⁶⁰Co spin direction → Parity violation by the weak interaction.

A. Gauge invariance → related to conservation of charge
B. Time invariance → energy
C. Translation invariance → momentum
D. Reflection invariance → parity
E. Rotation invariance → angular momentum

Answer: D

Special Relativity - Relativistic Energy

A lump of clay whose rest mass is 4 kilograms is travelling at three-fifths the speed of light when it collides head-on with an identical lump going the opposite direction at the same speed. If the two lumps stick together and no energy is radiated away, what is the mass of the composite lump?

A. 4 kg
B. 6.4 kg
C. 8 kg
D. 10 kg
E. 13.3 kg

(GR9677 #36)

Solution:

No energy is radiated away = energy is conserved

E_rel(a)+ E_rel(b)= E_rest(a,b)

γ_am_ac² + γ_bm_bc² = Mc²

Given:
m_a= m_b= m₀= 4 kg
|v_a|= |v_b|= ³/₅c

γ = 1/(1 − ^v²/_c²)^½
γ = 1/(1 − ⁹/₂₅)^½ = 1/(¹⁶/₂₅)^½ = 1/(⁴/5)^½= ⁵/₄
γ_a= γ_b= γ = ⁵/₄

γ_am_ac² + γ_bm_bc² = Mc²
M = 2γm₀ = 2(⁵/₄)(4) = 10 kg

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 - Energy

A ball is dropped from a height h. As it bounces off the floor, its speed is 80% of what it was just before it hit the floor. The ball will then rise to a height of most nearly

A. 0.94 h
B. 0.80 h
C. 0.75 h
D. 0.64 h
E. 0.50 h

(GR9277 #45)

Solution:

Conservation of energy before and when the ball hits the ground:

$\begin{aligned} \frac{1}{2}m \cdot 0+mgh&=\frac{1}{2}mv^2+mg\cdot 0 \\mh&=\frac{1}{2}v^2 \\v^2 &= 2gh \end{aligned}$

Conservation of energy when and after the ball hits the ground

$\begin{aligned} \frac{1}{2}mv'^2+mg\cdot 0&=\frac{1}{2}m\cdot 0+mgh' \\\frac{1}{2}mv'^2&=mgh' \\v'^2 &= 2gh' \\(0.8v)^2&=0.64v^2 = 0.64\times 2gh = 2gh' \\h'&=0.64h \end{aligned}$

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

Classical Mechanics - Rotational Motion

A hoop of mass M and radius R is at rest at the top of an inclined plane. The hoop rolls down the plane without slipping. When the hoop reached the bottom, its angular momentum around its center of mass is

A. MR√(gh)
B. ½ MR√(gh)
C. M√(2gh)
D. Mgh
E. ½ Mgh

(GR8677 #76)

Solution:

Conservation of energy: Total energy at the top = Total energy at the bottom

$Mgh= \frac{1}{2} Mv^2_{\text{CM}}+\frac{1}{2} I_\text{CM} \omega^2$

Velocity: v_CM = ωR
Moment inertia of the hoop: I_CM = MR²

$\\Mgh= \frac{1}{2} MR^2\omega^2+\frac{1}{2} MR^2 \omega^2=MR^2 \omega^2\\\\\rightarrow \omega^2 = \frac{gh}{R^2}\rightarrow \omega= \frac{\sqrt{gh}}{R}$

Angular Momentum: $L=I\omega=MR^2 \frac{\sqrt{gh}}{R} = MR\sqrt{gh}$

Answer: A

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

Nuclear & Particle Physics - The Sun’s Energy

The primary source of the sun’s energy is a series of thermonuclear reactions in which the energy produces is c² times the mass difference between

A. two hydrogen atoms and one helium atom
B. four hydrogen atoms and one helium atom
C. six hydrogen atoms and two helium atoms
D. three helium atoms and one carbon atom
E. two hydrogen atoms plus two helium atoms and one carbon atom

(GR0177 #19)

Solution:

In the Sun’s core, the protons of the hydrogen atoms (which largely make up the Sun) collide into each other and stick together or “fuse” to create helium nuclei.

4H → He + energy

It takes 4H nuclei to create a He nucleus due to the conservation of atomic mass.

Hydrogen atomic mass = 1 proton = 1.
Helium atomic mass = 2 protons + 2 neutrons = 4.

Answer: B

Classical Mechanics - Rotational Motion

A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point, as shown in the figure, and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?

A. $\sqrt{\frac{1}{3}gL}$
B. $\sqrt{gL}$
C. $\sqrt{3gL}$
D. $\sqrt{12gL}$
E. $12\sqrt{gL}$

(GR0177 #26)

Solution:

gr0177 #26b photo GR0177 26b_zpsib6ewhzg.png

Conservation of energy of the system before and after the rod falls onto the ground:
Mgy = ½Iω²

y = L/2 (center of the mass of the rod is at the center of the rod)
Moment inertia of uniform rod: I = ¹/₃ML²
v = rω
v_f= ωL → ω = v_f/L

Mg(L/2) = ½(¹/₃ML²)(v_f/L)²
gL = ¹/₃ v_f²
v_f = (3gL)^1/2

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

Nuclear & Particle Physics - Conservation Laws

The muon decays with a characteristic lifetime of about 10⁻⁶second into an electron, a muon neutrino, and an electron antineutrino. The muon is forbidden from decaying into an electron and just a single neutrino by the law of conservation of

A. charge
B. mass
C. energy and momentum
D. baryon number
E. lepton number

(GR0177 #78)

Solution:

If μ → e⁻+ ν

A. FALSE. Conservation of charge: − 1 → − 1 + 0 is not violated.

B. FALSE. Conservation of mass cannot be violated in any decay/interactions.

C. FALSE. Conservation of energy and momentum cannot be violated in any decay/interactions.

D. FALSE. Muon is a lepton. It follows the Conservation of Lepton Number, not Baryon Number.

E. TRUE. Lepton Numbers: 1 → 1 + 1 is violated.

Answer: E

Classical Mechanics - Rotational Motion

A child is standing on the edge of a merry-go-around that has the shape of a solid disk, as shown in the figure. The mass of the child is 40 kilograms. The merry-go-around has a mass 200 kilograms and a radius of 2.5 meters and it is rotating with an angular velocity of 2.0 radians per second. The child then walk slowly toward the center of the merry-go-around. What will the final angular velocity of the merry-go-around when the child reaches the center? (The size of the child can be neglected).

A. 2.0 rad/s
B. 2.2 rad/s
C. 2.4 rad/s
D. 2.6 rad/s
E. 2.8 rad/s

(GR0177 #89)

Solution:

Angular Momentum: L = Iω

Conservation of Angular Momentum:

∑ L_i = ∑ L_f

(I_m+ I_c)ω = (I_m'+ I_c')ω'
ω' = ω(I_m+ I_c)/(I_m'+ I_c')

Moment inertia of the uniform disk is the same before and after the child moves to the center:

I_m= I_m' = I_m= ½ m_mr²= ½ (200)(2.5)²
ω= 2 rad/s

Initally, moment inertia of the child:

I_c= m_cr²= (40)(2.5)²

When the child reaches the center, r= 0:

I_c' = 0

Thus,

ω' = ω(I_m+ I_c)/(I_m'+ I_c')
= (2) [½(200)(2.5)²+ (40)(2.5)²] / [½ (200)(2.5)²]
= (2) [½(200)+ (40)] / [½ (200)]
= 280/100
= 2.8 rad/s

Answer: E

Classical Mechanics - Rotational Motion

The cylinder shown above, with mass M and radius R, has a radially dependent density. The cylinder starts from from rest and rolls without slipping down an inclined plane of height H. At the bottom of the plane its translational speed is (8gH/7)^{1/2. Which of the following is the rotational inertia of the cylinder?}

$\\$A.$ \hspace{5 mm} \frac{1}{2}MR^2 \\\\$B.$ \hspace{5 mm} \frac{3}{4}MR^2 \\\\$C.$ \hspace{5 mm} \frac{7}{8}MR^2 \\\\$D.$ \hspace{5 mm} MR^2 \\\\$E.$ \hspace{5 mm} 4MR^2$

(GR0177 #91)

Solution:

Initally, it starts from rest, from height H,

At the bottom of the plane,

$\begin{aligned} E_f&=T_f+V_f=T_{\text{trans}}+T_{\text{rot}}+0 \\&=\frac{1}{2}Mv^2+\frac{1}{2}I \omega^2 \end{aligned}$

Conservation of Energy,

$\begin{aligned} E_i&=E_f \\MgH&=\frac{1}{2}Mv^2+\frac{1}{2}I \omega^2 \\I&=\frac{2MgH-Mv^2}{\omega^2} \end{aligned}$

with $v=R\omega$ and $(8gH/7)^{1/2}$

$\begin{aligned} I&=\frac{R^2}{v^2}\left( 2MgH-Mv^2 \right) \\&=\frac{2MgHR^2}{v^2}-MR^2 \\&=\frac{14MgHR^2}{8gH}-MR^2 \\&=\frac{7}{4}MR^2-MR^2 \\&=\frac{3}{4}MR^2 \end{aligned}$

Answer: B

Nuclear & Particle Physics – Conservation Laws

Which of the following reasons explains why a photon cannot decay to an electron and a positron γ → e⁻+ e⁺in free space?

A. Linear momentum and energy are not both conserved
B. Linear momentum and angular momentum are not both conserved
C. Angular momentum and parity are not both conserved
D. Parity and strangeness are not both conserved
E. Charge and lepton number are not both conserved

(GR9677 #96)

Solution:

See GR0177 #99

Answer: A

Subscribe to: Posts ( Atom )