Slide 1 / 42
Momentum
Free Response Problems
Slide 2 / 42
1. Block 1 with a mass of 500 g moves at a constant
speed of 5 m/s on a horizontal frictionless track and
collides and sticks to a stationary block 2 mass of 1.5
kg. Block 2 is attached to an unstretched spring with a
spring constant 200 N/m.
a. Determine the momentum of block 1 before the
collision.
b. Determine the kinetic energy of block 1 before the
collision.
c. Determine the momentum of the system of two
blocks after the collision.
d. Determine the velocity of the system of two blocks
after the collision.
e. Determine the kinetic energy of the system two
blocks after the collision.
f. Determine the maximum compression in the spring
after the collision.
Slide 3 / 42
1. Block 1 with a mass of 500 g moves at a constant
speed of 5 m/s on a horizontal frictionless track and
collides and sticks to a stationary block 2 mass of 1.5
kg. Block 2 is attached to an unstretched spring with a
spring constant 200 N/m.
a. Determine the momentum of block 1 before the
collision.
p1 = m1v1 = (0.5 kg)(5 m/s) = 2.5 kg m/s
Slide 4 / 42
1. Block 1 with a mass of 500 g moves at a constant
speed of 5 m/s on a horizontal frictionless track and
collides and sticks to a stationary block 2 mass of 1.5
kg. Block 2 is attached to an unstretched spring with a
spring constant 200 N/m.
b. Determine the kinetic energy of block 1 before the
collision.
KEo = ½m1v12 = ½(0.5 kg)(5 m/s)2 = 6.25 J
Slide 5 / 42
1. Block 1 with a mass of 500 g moves at a constant
speed of 5 m/s on a horizontal frictionless track and
collides and sticks to a stationary block 2 mass of 1.5
kg. Block 2 is attached to an unstretched spring with a
spring constant 200 N/m.
c. Determine the momentum of the system of two
blocks after the collision.
p1 + p2 = p'
p' = m1v1 + m2v2
p' = (0.5 kg)(5 m/s) = 2.5 kg m/s
Slide 6 / 42
1. Block 1 with a mass of 500 g moves at a constant
speed of 5 m/s on a horizontal frictionless track and
collides and sticks to a stationary block 2 mass of 1.5
kg. Block 2 is attached to an unstretched spring with a
spring constant 200 N/m.
d. Determine the velocity of the system of two blocks
after the collision.
p1 = (m1 + m2)v'
v' = p'/(m1 + m2)
v' = (2.5 kg m/s)/(0.5 kg + 1.5kg) = 1.25 m/s
Slide 7 / 42
1. Block 1 with a mass of 500 g moves at a constant
speed of 5 m/s on a horizontal frictionless track and
collides and sticks to a stationary block 2 mass of 1.5
kg. Block 2 is attached to an unstretched spring with a
spring constant 200 N/m.
e. Determine the kinetic energy of the system two
blocks after the collision.
KEF = ½Mv'2
KEF = ½(m1 + m2)(v')2
KEF = ½(0.5 kg + 1.5 kg)(1.25 m/s)2 = 1.56 J
Slide 8 / 42
1. Block 1 with a mass of 500 g moves at a constant
speed of 5 m/s on a horizontal frictionless track and
collides and sticks to a stationary block 2 mass of 1.5
kg. Block 2 is attached to an unstretched spring with a
spring constant 200 N/m.
f. Determine the maximum compression in the spring
after the collision.
Eo + W = Ef
KEo = EPEF
2
KEo = ½kx
x = (2KEo/k)1/2
x = [(2)(1.56 J)/(200 N/m)]
1/2
= 0.12 m
Slide 9 / 42
2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of
clay collides and sticks to a massive ball of mass 900 g suspended at the end
of a string.
a.
Calculate the momentum of the piece of clay
before the collision.
b.
Calculate the kinetic energy of the piece of
clay before the collision.
c.
What is the momentum of two objects after the collision?
d.
Calculate the velocity of the combination of two objects after the collision.
e.
Calculate the kinetic energy of the combination of two objects after the
collision.
f.
Calculate the change in kinetic energy during the collision.
g.
Calculate the maximum vertical height of the combination of two objects
after the collision.
Slide 10 / 42
2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of
clay collides and sticks to a massive ball of mass 900 g suspended at the end
of a string.
a.
Calculate the momentum of the piece of clay
before the collision.
p1 = m1v1 = (0.02 kg)(15 m/s) = 0.3 kg m/s
Slide 11 / 42
2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of
clay collides and sticks to a massive ball of mass 900 g suspended at the end
of a string.
b.
Calculate the kinetic energy of the piece of
clay before the collision.
KEo = ½(0.02 kg)(15 m/s)2 = 2.25 J
Slide 12 / 42
2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of
clay collides and sticks to a massive ball of mass 900 g suspended at the end
of a string.
c.
What is the momentum of two objects after
the collision?
p1 + p2 = p'
p' = p1 + p2
p' = m1v1 + m2v2
2
p' = (0.02 kg)(15 m/s) = 0.3 kg m/s
Slide 13 / 42
2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of
clay collides and sticks to a massive ball of mass 900 g suspended at the end
of a string.
d.
Calculate the velocity of the combination of two
objects after the collision.
p' = Mv'
p' = (m1 + m2)v'
v' = p'/(m1 + m2)
v' = (0.3 kg m/s)/(0.02 kg + 0.9 kg) = 0.33 m/s
Slide 14 / 42
2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of
clay collides and sticks to a massive ball of mass 900 g suspended at the end
of a string.
e.
Calculate the kinetic energy of the combination
of two objects after the collision.
KEF = ½(m1 + m2)v'
KEF = ½(0.02 kg + 0.9 kg)(0.33 m/s)2 = 0.05 J
Slide 15 / 42
2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of
clay collides and sticks to a massive ball of mass 900 g suspended at the end
of a string.
f.
Calculate the change in kinetic energy during
the collision.
# KE = KEF - KEo
# KE = (0.05 J) - (2.25 J) = -2.2 J
Slide 16 / 42
2. A 20 g piece of clay moves with a constant speed of 15 m/s. The piece of
clay collides and sticks to a massive ball of mass 900 g suspended at the end
of a string.
g.
Calculate the maximum vertical height of the
combination of two objects after the collision.
Eo + W = Ef
Eo = EF
KE = Mgh
h = KE/Mg
h = KE/(m1 + m2)g
2
h = (0.05 J)/(0.02 kg + 0.90 kg)(9.8 m/s ) = 0.006 m
Slide 17 / 42
3. A 10 g bullet moves at a constant speed
of 500 m/s and collides with a 1.5 kg
wooden block initially at rest. The surface of
the table is frictionless and 70 cm above the
floor level. After the collision the bullet
becomes embedded into the block. The
bullet-block system slides off the top of the
table and strikes the floor.
a.
Find the momentum of the bullet before the collision.
b.
Find the kinetic energy of the bullet before the collision.
c.
Find the velocity of the bullet-block system after the collision.
d.
Find the kinetic energy of the bullet-block after the collision.
e.
Find the change in kinetic energy during the collision.
f.
How much time it takes the bullet-block system to reach the floor?
g.
Find the maximum horizontal distance between the table and the striking
point on the floor.
Slide 18 / 42
3. A 10 g bullet moves at a constant speed
of 500 m/s and collides with a 1.5 kg
wooden block initially at rest. The surface of
the table is frictionless and 70 cm above the
floor level. After the collision the bullet
becomes embedded into the block. The
bullet-block system slides off the top of the
table and strikes the floor.
a.
Find the momentum of the bullet before the collision.
p1 = m1v1 = (0.01 kg)(500 m/s) = 5 kg m/s
Slide 19 / 42
3. A 10 g bullet moves at a constant speed
of 500 m/s and collides with a 1.5 kg
wooden block initially at rest. The surface of
the table is frictionless and 70 cm above the
floor level. After the collision the bullet
becomes embedded into the block. The
bullet-block system slides off the top of the
table and strikes the floor.
b.
Find the kinetic energy of the bullet before the collision.
KEo = KE1 + KE2
KEo = ½m1v12
KEo = ½(0.01 kg)(500 m/s)2
KEo = 1,250 J
Slide 20 / 42
3. A 10 g bullet moves at a constant speed
of 500 m/s and collides with a 1.5 kg
wooden block initially at rest. The surface of
the table is frictionless and 70 cm above the
floor level. After the collision the bullet
becomes embedded into the block. The
bullet-block system slides off the top of the
table and strikes the floor.
c.
Find the velocity of the bullet-block system after the collision.
p' = p1 + p2
(m1 + m2)v' = m1v1
v' = (m1v1)/(m1 + m2)
v' = (0.01 kg)(500 m/s)/(0.01 kg + 1.6 kg)
v' = 3.3 m/s
Slide 21 / 42
3. A 10 g bullet moves at a constant speed
of 500 m/s and collides with a 1.5 kg
wooden block initially at rest. The surface of
the table is frictionless and 70 cm above the
floor level. After the collision the bullet
becomes embedded into the block. The
bullet-block system slides off the top of the
table and strikes the floor.
d.
Find the kinetic energy of the bullet-block after the collision.
KEF = ½(m1 + m2)v'2
KEF = ½(0.01 kg + 1.5 kg)(3.3 m/s)2
KEF = 8.2 J
Slide 22 / 42
3. A 10 g bullet moves at a constant speed
of 500 m/s and collides with a 1.5 kg
wooden block initially at rest. The surface of
the table is frictionless and 70 cm above the
floor level. After the collision the bullet
becomes embedded into the block. The
bullet-block system slides off the top of the
table and strikes the floor.
e.
Find the change in kinetic energy during the collision.
ΔKE = KEF - KEo
ΔKE = 8.2 J - 1250 J
ΔKE = -1241.8 J
Slide 23 / 42
3. A 10 g bullet moves at a constant speed
of 500 m/s and collides with a 1.5 kg
wooden block initially at rest. The surface of
the table is frictionless and 70 cm above the
floor level. After the collision the bullet
becomes embedded into the block. The
bullet-block system slides off the top of the
table and strikes the floor.
f.
How much time it takes the bullet-block system to reach the floor?
y = yo + vot + ½at2
y = ½at2
t = (2y/g)1/2
2
t = [(2)(0.7 m)/(9.8 m/s )]
t = 0.38 s
1/2
Slide 24 / 42
3. A 10 g bullet moves at a constant speed
of 500 m/s and collides with a 1.5 kg
wooden block initially at rest. The surface of
the table is frictionless and 70 cm above the
floor level. After the collision the bullet
becomes embedded into the block. The
bullet-block system slides off the top of the
table and strikes the floor.
g.
Find the maximum horizontal distance between the table and the striking
point on the floor.
Δx = vxt
Δx = (3.3 m/s)(0.38s)
Δx = 1.25 m
Slide 25 / 42
4. Block A with a mass of m is released
from the top of the curved track of radius
r. Block A slides down the track without
friction and collides inelastically with an
identical block B initially at rest. After the
collision the two blocks move distance X
to the right on the rough horizontal part
of the track with a coefficient of kinetic
friction µ.
a.
What is the speed of block A just before it hits block B?
b.
What is the speed of the system of two blocks after the collision?
c.
What is the kinetic energy of the system of two blocks the
collision?
d.
How much energy is lost due to the collision?
e.
What is the stopping distance X of the system of two blocks?
Slide 26 / 42
4. Block A with a mass of m is released
from the top of the curved track of radius
r. Block A slides down the track without
friction and collides inelastically with an
identical block B initially at rest. After the
collision the two blocks move distance X
to the right on the rough horizontal part
of the track with a coefficient of kinetic
friction µ.
a.
What is the speed of block A just before it hits block B?
Eo + W = EF
mgh = ½mv2
v = (2gh)1/2
v = (2gr)1/2
Slide 27 / 42
4. Block A with a mass of m is released
from the top of the curved track of radius
r. Block A slides down the track without
friction and collides inelastically with an
identical block B initially at rest. After the
collision the two blocks move distance X
to the right on the rough horizontal part
of the track with a coefficient of kinetic
friction µ.
b.
What is the speed of the system of two blocks after the collision?
p = p'
mv + 0 = v'(m + m)
mv = 2mv'
v' = v/2
v' = [(2gr)1/2]/2
since v = (2gr)1/2,
Slide 28 / 42
4. Block A with a mass of m is released
from the top of the curved track of radius
r. Block A slides down the track without
friction and collides inelastically with an
identical block B initially at rest. After the
collision the two blocks move distance X
to the right on the rough horizontal part
of the track with a coefficient of kinetic
friction µ.
c.
What is the kinetic energy of the system of two blocks after the
collision?
KE' = ½(m + m)v'2
KE' = ½(2m)[(2gr)1/2/2]2
KE' = 2grm/4
KE' = grm/2
Slide 29 / 42
4. Block A with a mass of m is released
from the top of the curved track of radius
r. Block A slides down the track without
friction and collides inelastically with an
identical block B initially at rest. After the
collision the two blocks move distance X
to the right on the rough horizontal part
of the track with a coefficient of kinetic
friction µ.
d.
How much energy is lost due to the collision?
ΔE = EF - Eo
ΔE = mgr/2 - mgr
ΔE = -mgr/2 (energy lost)
Slide 30 / 42
4. Block A with a mass of m is released
from the top of the curved track of radius
r. Block A slides down the track without
friction and collides inelastically with an
identical block B initially at rest. After the
collision the two blocks move distance X
to the right on the rough horizontal part
of the track with a coefficient of kinetic
friction µ.
e.
What is the stopping distance X of the system of two blocks?
W = F# X
ΔX = W/F
ΔX = (mgr/2)/µmg
ΔX = r/2µ
Slide 31 / 42
5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal
frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x
direction and disc m2 is initially at rest. The collision of two discs is perfectly
elastic and the directions of two velocities presented by the diagram.
a.
What is the x-component of the initial momentum of disc m1?
b.
What is the y-component of the initial momentum of disc m1?
c.
What is the x-component of the initial momentum of disc m2?
d.
What is the y-component of the initial momentum of disc m2?
e.
What is the x-component of the final momentum of disc m1?
f.
What is the x-component of the final momentum of disc m2?
g.
What is the y-component of the final momentum of disc m2?
h.
What is the final vector velocity of m2?
i.
What is the y-component of the final momentum of disc m1?
j.
What is the final vector velocity of disc m1?
Slide 32 / 42
5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal
frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x
direction and disc m2 is initially at rest. The collision of two discs is perfectly
elastic and the directions of two velocities presented by the diagram.
a.
What is the x-component of the initial momentum of disc m1?
p = (2 kg)(8 m/s)
p = 16 kg m/s
Slide 33 / 42
5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal
frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x
direction and disc m2 is initially at rest. The collision of two discs is perfectly
elastic and the directions of two velocities presented by the diagram.
b.
0
What is the y-component of the initial momentum of disc m1?
Slide 34 / 42
5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal
frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x
direction and disc m2 is initially at rest. The collision of two discs is perfectly
elastic and the directions of two velocities presented by the diagram.
c.
0
What is the x-component of the initial momentum of disc m2?
Slide 35 / 42
5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal
frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x
direction and disc m2 is initially at rest. The collision of two discs is perfectly
elastic and the directions of two velocities presented by the diagram.
d.
0
What is the y-component of the initial momentum of disc m2?
Slide 36 / 42
5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal
frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x
direction and disc m2 is initially at rest. The collision of two discs is perfectly
elastic and the directions of two velocities presented by the diagram.
e.
0
What is the x-component of the final momentum of disc m1?
Slide 37 / 42
5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal
frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x
direction and disc m2 is initially at rest. The collision of two discs is perfectly
elastic and the directions of two velocities presented by the diagram.
f.
What is the x-component of the final momentum of disc m2?
px + I = px'
p2'
30
o
p2x'
p2y'
p1x + 0 = 0 + p2x'
p2x' = p1x = 16 kg m/s
Slide 38 / 42
5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal
frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x
direction and disc m2 is initially at rest. The collision of two discs is perfectly
elastic and the directions of two velocities presented by the diagram.
g.
What is the y-component of the final momentum of disc m2?
tan 30o = p2y'/p2x'
p2'
30
o
16 kg m/s
p2y'
p2y' = p2x'(tan30o)
p2y' = (16 kg m/s)(tan 30o)
p2y' = 9.2 kg m/s
Slide 39 / 42
5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal
frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x
direction and disc m2 is initially at rest. The collision of two discs is perfectly
elastic and the directions of two velocities presented by the diagram.
h.
What is the final vector velocity of m2?
p2'
30
o
16 kg m/s
9.2 kg m/s
p2' = (p2x'2 + p2y'2)1/2
p2' = [(16 kg m/s)2 + (9.2 kg m/s)2]1/2
p2' = 18.5 kg m/s
p2' = m2v2'
v2' = p2'/m2
v2' = (18.5 kg m/s)/(8 kg)
v2' = 2.3 m/s
Slide 40 / 42
5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal
frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x
direction and disc m2 is initially at rest. The collision of two discs is perfectly
elastic and the directions of two velocities presented by the diagram.
i.
What is the y-component of the final momentum of disc m1?
py + I = py'
0 = p1y' + p2y'
p1y' = -p2y'
p1y' = -9.2 kg m/s
Slide 41 / 42
5. Two discs of masses m1 = 2kg and m2 = 8 kg are placed on a horizontal
frictionless surface. Disc m1 moves at a constant speed of 8 m/s in +x
direction and disc m2 is initially at rest. The collision of two discs is perfectly
elastic and the directions of two velocities presented by the diagram.
j.
What is the final vector velocity of disc m1?
p1' = m1v1'
v1' = p1'/m1
v1' = (-9.2 kg m/s)/(2 kg)
v1' = -4.6 m/s
Slide 42 / 42