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 - 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_A v_A + m_B v_B = m_A v_A' + m_B v_B'
mv + 0 = 0 + MV
V = m/Mv

Answer: A