Problem#1
Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 13.0 m/s toward the east, and the other is traveling north with speed v2i. Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of 55.0° north of east. The speed limit for both roads is 35 mi/h, and the driver of the northward-moving vehicle claims he was within the speed limit when the collision occurred. Is he telling the truth?Answer:
We use conservation of momentum for the system of two vehicles for both northward and eastward components.
For the eastward direction:
m(13.0 m/s) + 0 = 2mvfcos55.00
vf = 11.3 m/s
For the northward direction:
mv2i = 2mvfsin55.00
v2i = 2(11.3 m/s)sin55.00 = 18.6 m/s = 41.5 mi/h
Thus, the driver of the north bound car was untruthful.
Problem#2
A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves, at 4.33 m/s, at an angle of 30.0° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball’s velocity after the collision.
Answer:
By conservation of momentum for the system of the two billiard balls (with all masses equal),
The x-axis:
m1(5.00 m/s) + 0 = m1(4.33 m/s)cos30.00 + m2v2f cosφ
v2f cosφ = 5.00 m/s – (4.33 m/s)cos30.00 = 1.25 m/s (1)
The y-axis:
0 = m1(4.33 m/s)sin30.00 + m2v2f sinφ
v2f sinφ = –2.17 m/s
from (1) and (2) we get
tanφ = –1.732 or φ = –60.00
so, v2f = 2.50 m/s at –60.00
Note that we did not need to use the fact that the collision is perfectly elastic.
Problem#3
A proton, moving with a velocity of vii, collides elastically with another proton that is initially at rest. If the two protons have equal speeds after the collision, find (a) the speed of each proton after the collision in terms of vi and (b) the direction of the velocity vectors after the collision.
Answer:
(a)
(a) By conservation of momentum of the system:
m1v1 + m2v2 + m3v3 = 0
(5.00 x 10-27 kg)(6.00 x 106 m/s)j + (8.40 x 10-27 kg)(4.00 x 106 m/s)i + (3.60 x 10-27 kg)v3 = 0
v3 = –(4.00i + 8.33 j) x 106 m/s
(b) the total kinetic energy increase in the process given by
∆K = K1 + K2 + K3
∆K = ½(5.00 x 10-27 kg)(6.00 x 106 m/s)2 + ½(8.40 x 10-27 kg)(4.00 x 106 m/s)2 + ½(3.60 x 10-27 kg)[(4.00 x 106 m/s)2 + (8.33 x 106 m/s)2]
∆K = 4.39 x 10-13 J = 2.74 MeV
The x-axis:
m1(5.00 m/s) + 0 = m1(4.33 m/s)cos30.00 + m2v2f cosφ
v2f cosφ = 5.00 m/s – (4.33 m/s)cos30.00 = 1.25 m/s (1)
The y-axis:
0 = m1(4.33 m/s)sin30.00 + m2v2f sinφ
v2f sinφ = –2.17 m/s
from (1) and (2) we get
tanφ = –1.732 or φ = –60.00
so, v2f = 2.50 m/s at –60.00
Note that we did not need to use the fact that the collision is perfectly elastic.
Problem#3
A proton, moving with a velocity of vii, collides elastically with another proton that is initially at rest. If the two protons have equal speeds after the collision, find (a) the speed of each proton after the collision in terms of vi and (b) the direction of the velocity vectors after the collision.
Answer:
(a)
We use the law of conservation of momentum, pi = pf, then
pxi = pxf
mvi + 0 = mv cosθ+ mvcosφ (1)
and pyi = pyf
0 = mv sinθ+ mvsinφ (2)
sinθ = –sinφ or θ = –φ
Furthermore, energy conservation for the system of two protons requires
½ mvi2 = ½ mv2 + ½ mv2
v = vi/√2
(b) Hence, (1) gives
mvi + 0 = m(vi/√2) cosθ+ m(vi/√2)cos(–θ)
√2 = 2cosθ
θ = 45.00, then φ = –45.00
Problem#4
Two particles with masses m and 3m are moving toward each other along the x axis with the same initial speeds vi. Particle m is traveling to the left, while particle 3m is traveling to the right. They undergo an elastic glancing collision such that particle m is moving downward after the collision
at right angles from its initial direction. (a) Find the final speeds of the two particles. (b) What is the angle θ at which the particle 3m is scattered?
Answer:
(a) x-component of momentum for the system of the two objects:
p1ix + p2ix = p1fx + p2fx
–mvi + 3mvi = 0 + 3mv2x
v2x = 2vi/3
y-component of momentum of the system:
0 + 0 = –mv1y + 3mv2y
v1y = 3v2y
by conservation of energy of the system:
½ mvi2 + ½ (3m)vi2 = ½ mv1y2 + ½ (3m)[v2x2 + v2y2]
4vi2 = (3v2y)2 + (3m)[(2vi/3)2 + v2y2]
8vi2/3 = 12v2y
v2y = vi√2/3
The object of mass m has final speed from (2)
v1y = vi√2
and the object of mass 3 m moves at
[(v2x)2 + (v2y)2]1/2 = [(2vi/3)2 + (√2vi/3)]1/2
= vi√6/3
(b) the angle θ at which the particle 3m is scattered is
θ = tan-1(v2y/v2x) = tan-1[(vi√2/3)(3/2vi)] = 35.30
Problem#5
An unstable atomic nucleus of mass 17.0 x 10-27 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.00 x 10-27 kg, moves along the y axis with a speed of 6.00 x 106 m/s. Another particle, of mass 8.40 x 10-27 kg, moves along the x axis with a speed of 4.00 x 106 m/s. Find (a) the velocity of the third particle and (b) the total kinetic energy increase in the process.
Answer:
Given: m0 = 17.0 x 10-27 kg, m1 = 5.00 x 10-27 kg, m2 = 8.40 x 10-27 kg and
m3 = m0 – m1 – m2 = 3.60 x 10-27 kg
v1 = (6.00 x 106 m/s)j
v2 = (4.00 x 106 m/s)i
pxi = pxf
mvi + 0 = mv cosθ+ mvcosφ (1)
and pyi = pyf
0 = mv sinθ+ mvsinφ (2)
sinθ = –sinφ or θ = –φ
Furthermore, energy conservation for the system of two protons requires
½ mvi2 = ½ mv2 + ½ mv2
v = vi/√2
(b) Hence, (1) gives
mvi + 0 = m(vi/√2) cosθ+ m(vi/√2)cos(–θ)
√2 = 2cosθ
θ = 45.00, then φ = –45.00
Problem#4
Two particles with masses m and 3m are moving toward each other along the x axis with the same initial speeds vi. Particle m is traveling to the left, while particle 3m is traveling to the right. They undergo an elastic glancing collision such that particle m is moving downward after the collision
at right angles from its initial direction. (a) Find the final speeds of the two particles. (b) What is the angle θ at which the particle 3m is scattered?
Answer:
(a) x-component of momentum for the system of the two objects:
p1ix + p2ix = p1fx + p2fx
–mvi + 3mvi = 0 + 3mv2x
v2x = 2vi/3
y-component of momentum of the system:
0 + 0 = –mv1y + 3mv2y
v1y = 3v2y
by conservation of energy of the system:
½ mvi2 + ½ (3m)vi2 = ½ mv1y2 + ½ (3m)[v2x2 + v2y2]
4vi2 = (3v2y)2 + (3m)[(2vi/3)2 + v2y2]
8vi2/3 = 12v2y
v2y = vi√2/3
The object of mass m has final speed from (2)
v1y = vi√2
and the object of mass 3 m moves at
[(v2x)2 + (v2y)2]1/2 = [(2vi/3)2 + (√2vi/3)]1/2
= vi√6/3
(b) the angle θ at which the particle 3m is scattered is
θ = tan-1(v2y/v2x) = tan-1[(vi√2/3)(3/2vi)] = 35.30
Problem#5
An unstable atomic nucleus of mass 17.0 x 10-27 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.00 x 10-27 kg, moves along the y axis with a speed of 6.00 x 106 m/s. Another particle, of mass 8.40 x 10-27 kg, moves along the x axis with a speed of 4.00 x 106 m/s. Find (a) the velocity of the third particle and (b) the total kinetic energy increase in the process.
Answer:
Given: m0 = 17.0 x 10-27 kg, m1 = 5.00 x 10-27 kg, m2 = 8.40 x 10-27 kg and
m3 = m0 – m1 – m2 = 3.60 x 10-27 kg
v1 = (6.00 x 106 m/s)j
v2 = (4.00 x 106 m/s)i
(a) By conservation of momentum of the system:
m1v1 + m2v2 + m3v3 = 0
(5.00 x 10-27 kg)(6.00 x 106 m/s)j + (8.40 x 10-27 kg)(4.00 x 106 m/s)i + (3.60 x 10-27 kg)v3 = 0
v3 = –(4.00i + 8.33 j) x 106 m/s
(b) the total kinetic energy increase in the process given by
∆K = K1 + K2 + K3
∆K = ½(5.00 x 10-27 kg)(6.00 x 106 m/s)2 + ½(8.40 x 10-27 kg)(4.00 x 106 m/s)2 + ½(3.60 x 10-27 kg)[(4.00 x 106 m/s)2 + (8.33 x 106 m/s)2]
∆K = 4.39 x 10-13 J = 2.74 MeV
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