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CHAPTER CHAPTER 77 LINEAR LINEAR MOMENTUM MOMENTUM ,, IMPULSE IMPULSE AND AND COLLISIONS COLLISIONS
Dr. Abdallah M. Azzeer
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Linear Momentum The principle of energy conservation can be used to solve problems that are harder to solve just using Newton’s laws. It is used to describe motion of an object or a system of objects. A new concept of linear momentum can also be used to solve physical problems, especially the problems involving collisions of objects. Linear momentum of an object whose mass is m and is moving at a velocity of v is defined as
p = mv
Dr. Abdallah M. Azzeer
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What can you tell from this definition about momentum? Momentum is a vector quantity. The heavier the object the higher the momentum The higher the velocity the higher the momentum Its unit is kg.m/s What else can use see from the definition? Do you see force? The change of momentum in a given time interval vi
vf
F
fk = 0
Δt
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Dr. Abdallah M. Azzeer
a =
vf -v Δt
i
vf -vi ) Δt − mv i = pf − pi = Δp
F = ma = m( F Δt = m v f
Impulse = force times time=change in momentum
r r r I = F Δt = Δ p
G G JG G G G G JG Δ p mv − mv 0 m v − v 0 Δv = = =m = ma = ∑ F Δt Δt Δt Δt
(
)
READ Examples 7.1 & 7.2
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Momentum Conservation r ΔP =0 Δt
r r ΔP FEXT = Δt
r FEXT = 0
¾ The concept of momentum conservation is one of the most fundamental principles in physics. ¾ This is a component (vector) equation. - We can apply it to any direction in which there is no external force applied. ¾ You will see that we often have momentum conservation even when energy is not conserved.
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Dr. Abdallah M. Azzeer
Impulse and Momentum
A classic photograph taken by Arthur Edgerton at MIT in 1935 showing the moment of impact between bat and softball. The huge force exerted by the bat on the ball causes severe distortion of the ball as it is hit.
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Impulse and Momentum •
Newton’s 2nd Law:
r Δ r F= ( p) Δt
r r Δp = F Δt r r r r Δp = pf − pi = F Δt A classic photograph taken by Arthur Edgerton at MIT in 1935 showing the moment of impact between bat and softball. The huge force exerted by the bat on the ball causes severe distortion of the ball as it is hit.
r r I = F Δt
Impulse
ÆImpulse-momentum theorem
Dr. Abdallah M. Azzeer
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Impulse and Momentum r tf r I = ∫ Fdt ti
r
r
In general, F = F ( t )
r
ÆImpulse = area under F, t curve
r 1 tf r F = ∫ Fdt Δt ti r r I = FΔt Simple case: constant Force
r r I = FΔt
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Example; (a) Calculate the impulse experienced when a 70 kg person lands on firm ground after jumping from a height of 3.0 m. Then estimate the average force exerted on the person’s feet by the ground, if the landing is (b) stiff-legged and (c) with bent legs. In the former case, assume the body moves 1.0cm during the impact, and in the second case, when the legs are bent, about 50 cm. We don’t know the force. How do we do this? Obtain velocity of the person before striking the ground.
K E = − Δ PE
1 2 mv = − mg ( y − yi ) = mgyi 2
Solving the above for velocity v, we obtain
v = 2 gyi = 2 ⋅ 9.8 ⋅ 3 = 7.7 m / s Then as the person strikes the ground, the momentum becomes 0 quickly giving the impulse
I = F Δt = Δp = p f − pi = 0 − mv = = −70 kg ⋅ 7.7 m / s = −540 N ⋅ s Dr. Abdallah M. Azzeer
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Example; cont’d In coming to rest, the body decelerates from 7.7m/s to 0m/s in a distance d=1.0cm=0.01m.
0 + vi 7.7 = = 3.8 m / s 2 2 d 0.01m The time period the collision lasts is = 2.6 × 10 − 3 s Δt = = 3.8 m / s v I = F Δt = 540 N ⋅ s Since the magnitude of impulse is I 540 The average force on the feet during = = 2.1× 105 N F= this landing is Δt 2.6 × 10 −3 2 2 How large is this average force? Weight = 70 kg ⋅ 9.8m / s = 6.9 × 10 N The average speed during this period is
v=
F = 2.1× 105 N = 304 × 6.9 × 10 2 N = 304 × Weight If landed in stiff legged, the feet must sustain 300 times the body weight. The person will likely break his leg.
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Example ; cont’d What if the knees are bent in coming to rest? The body decelerates from 7.7 m/s to 0 m/s in a distance d=50 cm=0.5 m. The average speed during this period is still the same v =
0 + vi 7.7 = = 3.8 m / s 2 2
d 0.5 m = = 1.3 × 10 − 1 s 3.8 m / s v I = F Δt = 540 N ⋅ s Since the magnitude of impulse is I 540 The average force on the feet during = = 4.1× 103 N F= −1 this landing is Δt 1.3 × 10 2 2 How large is this average force? Weight = 70 kg ⋅ 9.8m / s = 6.9 × 10 N The time period the collision lasts changes to
Δt =
F = 4.1× 103 N = 5.9 × 6.9 × 10 2 N = 5.9 × Weight It’s only 6 times the weight that the feet have to sustain! So by bending the knee you increase the time of collision, reducing the average force exerted on the knee, and will avoid injury! 11
Dr. Abdallah M. Azzeer
Example for Impulse In a crash test, an automobile of mass 1500 kg collides with a wall. The initial and final velocities of the automobile are vi=-15.0 m/s and vf=2.60 m/s in the x- direction. If the collision lasts for 0.150 seconds, what would be the impulse caused by the collision and the average force exerted on the automobile? Let’s assume that the force involved in the collision is a lot larger than any other forces in the system during the collision. From the problem, the initial and final momentum of the automobile before and after the collision is
p i = m v i = 1500 × ( − 15.0 ) = − 22500 k g ⋅ m / s
p f = mv f = 1500 × ( 2.60 ) = 3900 k g ⋅ m / s Therefore the impulse on the I = Δ p = p − p i = ( 3900 + 22500 ) kg ⋅ m / s automobile due to the collision is = 26400 kg ⋅ m / s = 2.64 × 10 4 kg ⋅ m / s f
The average force exerted on the automobile during the collision is
2 . 64 × 10 Δ p F = Δt = 0 . 150
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= 1.76 ×105 N
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MOMENTUM CONSERVATION Consider two objects, 1 and 2, and assume that no external forces are acting on the system composed of these two particles. v2
v1
Impulse applied to object 1
r r r F21Δt = m1v 1 − m1u1
m1
m2
Before collision
Impulse applied to object 2
r r r F12 Δt = m 2v 2 − m 2u 2
m F21 1
Apply Newton’s Third Law
r r F12 = − F21 r r or F12 Δt = − F21Δt
m2 F12
During collision u2
u1 m1
m2
After collision
r r r r 0 = m1v1 − m1u1 + m2 v2 − m2u 2 r r r r m 1v 1 + m 2v 2 = m 1u 1 + m 2u 2
Total impulse applied to system
or
Dr. Abdallah M. Azzeer
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¨Net momentum of the two objects before and after collision is the same ¨ THE TOTAL MOMENTUM OF THE SYSTEM IS CONSERVED ¨ For conservation of momentum, the external forces must be zero In one dimension in component form,
m 1v 1x + m 2v 2 x = m 1u 1x + m 2u 2 x
m1v 1y + m 2v 2 y = m1u1y + m 2u 2 y READ Examples 7.3- 7.6
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Example: Figure Figure below below shows shows aa 2.0 2.0 kg kg toy toy race race car car before before and and after after taking taking aa turn turn on on aa track. track. Its Itsrspeed speed isis 0.50 0.50 m/s m/s before before the the turn turn and and 0.40 0.40 m/s m/s after after the the turn. turn. What What isis the the change change Δ Δ P in in the the linear linear momentum momentum of of the the car car due due to to the the turn? turn? r v i = − (0.50 m / s ) jˆ and
r v f = (0.40 m / s ) iˆ
r r Pi = M v i = (2.0 kg )( −0.50 m / s ) jˆ = ( −1 kg ⋅ m / s ) jˆ r r Pf = M v f = (2.0 kg )(0.40 m / s ) iˆ = (0.80 kg ⋅ m / s ) iˆ r r r Δ P = Pf − Pi
r Δ P = (0.80 kg ⋅ m / s ) iˆ − ( −1.0kg ⋅ m / s ) jˆ = (0.8 iˆ + 1.0 jˆ ) kg ⋅ m / s
Dr. Abdallah M. Azzeer
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Sample Problem AA pitched pitched 140 140 gg baseball, baseball, in in horizontal horizontal flight flight with with aa speed speed vvii of of 39.0 39.0 m/s, m/s, isis struck struck by by aa bat. bat. After After leaving leaving the the bat, bat, the the ball ball travels travels in in the the opposite opposite direction direction with with speed speed vvff ,, also also 39.0 39.0 m/s. m/s. (a) (a) What What impulse impulse II acts acts on on the the ball ball while while itit isis in in contact contact with with the the bat bat during during the the collision? collision? I = p f − p i = mv f − mv i = (0.140 kg )(39.0 m / s ) − (0.140 kg )( − 39.0 m / s ) = 10.9 kg ⋅ m / s
Note that I is the impulse on the ball. The final direction of the ball is positive. (b) The impact time Δt for the baseball-bat collision is 1.20 ms. What average force acts on the baseball? Favg =
I 10.9 kg ⋅ m / s = = 9080 N 0.00120 s Δt
Note that this average force is from the bat to the ball. The positive direction of the force is in the final velocity of the ball.
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(c) Now suppose the collision is not head-on, and the ball leaves the bat with a speed vf of 45.0 m/s at an upward angle of 30.0°. What now is the impulse on the ball? The impulse on the ball is : r r r r r r I = Δ p = p f − p i = mv f − mv i
I x = p f x − p i x = m (v f x − v i x ) = (0.140 kg ) [(45.0 m / s )(cos 30.0o ) − (− 39.0 m / s )] = 10.92 kg ⋅ m / s
I y = pf
y
− p i y = m (v f
y
−v i y )
= (0.140 kg ) [(45.0 m / s )(sin 30.0o ) − 0] = 3.150 kg ⋅ m / s
r I = (10.9 iˆ + 3.15 jˆ ) kg ⋅ m / s I = I x 2 + I y 2 = 11.4 kg ⋅ m / s
θ = tan −1
Iy Ix
= 16o
Dr. Abdallah M. Azzeer
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Sample Problem AA ballot ballot box box with with mass mass m m == 6.0 6.0 kg kg slides slides with with speed speed vv == 4.0 4.0 m/s m/s across across aa frictionless frictionless floor floor in in the the positive positive direction direction of of an an xx axis. axis. ItIt suddenly suddenly explodes explodes into into two two pieces. pieces. One One piece, piece, with with mass mass m m11 == 2.0 2.0 kg, kg, moves moves in in the the positive positive direction direction of of the the xx axis axis with with speed speed vv11 == 8.0 8.0 m/s. m/s. What What isis the the velocity velocity of of the the second second piece, piece, with with mass mass m m22 ?? r r Pi = mv r r r r Pf 1 = m 1v 1 and Pf 2 = m 2v 2 r r r r r Pf = Pf 1 + Pf 2 = m 1v 1 + m 2v 2
Pi = Pf mv = m 1v 1 + m 2v 2 (6.0 kg )(4.0 m / s ) = (2.0 kg )(8.0 m / s ) + (4.0 kg ) v 2 v 2 = 2.0 m / s
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Examples 7.3 AA neutron neutron moving moving at at 2700 2700 m/s m/s collides collides head head on on with with aa nitrogen nitrogen nucleus nucleus at at rest rest and and isis -27 -27 kg and M = absorbed. The neutron and nitrogen masses are m = 1.67 × 10 23.0 absorbed. The neutron and nitrogen masses are m = 1.67 × 10 kg - and M = 23.0 -27 kg, ×10 ×10-27 kg, respectively. respectively. What What isis the the final final velocity velocity of of the the combined combined object? object? M m V=0
v
u
Neutron Nitrogen Atom BEFORE
AFTER
Momentum before; pi = mv Momentum after;
pf = (m+M)u
Net force = 0 ¨
mv = (m+M)u Ö u =
mv = 183 m/s (m + M )
Whenever there is no net external force acting on a system. its momentum is conserved. Momentum is therefore always conserved for an isolated system, one subjected only to internal forces. Dr. Abdallah M. Azzeer
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Examples 7.4 A cannon is mounted inside a railroad car, which is initially at rest but can move frictionlessly (Fig. 7.5). It fires a cannonball of mass m = 5 kg with a horizontal velocity v = 15 m/s relative to the ground at the opposite wall. The total mass of the cannon and railroad car is M = 15,000 kg. (Assume that the mass of the exhaust gases is negligible.) (a) What is the velocity V of the car while the cannonball is in flight? (b) If the cannonball becomes embedded in the wall, what is the velocity of the car and ball after impact?
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(a) Momentum before; pi = 0 Momentum after; pf = mv+MV Net force = 0
m v = −5 × 10−3 m/s M The recoil speed of the car and cannon is very small because of their large mass.
¨
0 = mv+MV
Ö
V =−
(b) As the ball becomes embedded in the wall, it exerts a force on the wall to the right in Fig. 7.5. The wall, in turn, exerts a force to the left on the ball. The ball and car both stop moving when this happens, since the net momentum is still zero. Meanwhile the car will have rolled to the left as the ball travelled to the right.
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Dr. Abdallah M. Azzeer
Examples 7.5 The ballistic pendulum uses conservation laws to find the velocity of a bullet
AA gun gun isis fired fired horizontally horizontally into into aa wooden wooden block block suspended suspended by by strings strings (Fig. (Fig. 7.6). 7.6). The The bullet bullet stops stops in in the the block, block, which which rises rises 0.2 0.2 m. m. The The mass mass of of the the bullet bullet isis 0.03 0.03 kg, kg, and and the the mass mass of of the the block block isis 22 kg. kg. (a) (a) What What was was the the velocity velocity of of the the block block just just after after the the bullet bullet stopped stopped in in it? it? (b) (b) What What was was the the velocity velocity of of the the bullet bullet before before itit struck struck the the block? block?
V=0 m
r v
h
r V
M
Dr. Abdallah M. Azzeer
M+m
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Two stage process: The first is the rapid "collision" of the bullet and the block. The second is the subsequent rise of the block plus the embedded bullet. Stage 1: Momentum is conserved
mv = (m + M ) V
⎛ m ⎞ V =⎜ ⎟v ⎝m +M ⎠
Stage 2: K+U Energy is conserved
(Ei = Ef ) 1 2
(m + M ) V 2 = (m + M )gh
Eliminating V gives:
V
2
= 2 gh
V = 2gh = 1.98 m/s
⎛ M⎞ v = ⎜ 1 + ⎟ 2gh = 134 m/s = 482 km/hr ⎝ m⎠
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Dr. Abdallah M. Azzeer
It is important to realize that momentum is a vector quantity, and that if the total momentum of a system is constant, each component must be constant.
Examples 7.6 AA car car of of mass mass m m == 1000 1000 kg kg moving moving at at 30 30 m/s m/s collides collides with with aa car car of of mass mass M M == 2000 2000 kg kg travelling travelling at at 20 20 m/s m/s in in the the opposite opposite direction. direction. Immediately Immediately after after the the collision, collision, the the 1000-kg 1000-kg car car moves moves at at right right angles angles to to its its original original direction direction at at 15 15 m/s. m/s. Find Find the the velocity velocity of of the the 2000-kg 2000-kg car car right right after after the the collision collision Vf y
y Vi x m
vi
M x
vf
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Conservation of momentum: r r r r mv i + MV i = mv f + MV f (1)
Split into components: mv ix + MV ix = mv fx + MV fx (2) mv iy + MV iy = mv fy + MV fy
(3)
Where ; vix= +30 m/s, vfx = 0, Vix= -20 m/s viy = 0, Viy = 0, vfy = - 15 m/s, Vfy = ? From (2) ¨ V fx =
mv ix + MV ix = −5 m/s M
From (3) ¨ V fy = −
Vf θ
y
m v = +7.5 m/s M fy
MORE
x
V f = V fx2 +V fy2 = 9 m/s
θ =tan -1 (V fy /V fx ) = 56.3o vf
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7.47.4- Elastic and Inelastic COLLISIONS Collisions Collisions involve involve forces forces internal internal to to colliding colliding bodies. bodies. ¾ ¾ Perfectly Perfectly elastic elastic collisions collisions -- conserve conserve energy energy and and momentum momentum ¾ ¾ Inelastic Inelastic collisions collisions -- conserve conserve momentum momentum
Totally Totally inelastic inelastic collisions collisions -- conserve conserve momentum momentum and and objects objects stick stick together together
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اﻟﺘﺼﺎدم اﻟﻤﺮن Elastic collision ﻓﻰ هﺬا اﻟﻨﻮع ﺗﻜﻮن آﻤﻴﺔ اﻟﺤﺮآﺔ momentumو اﻟﻄﺎﻗﺔ اﻟﺤﺮآﻴﺔ kinetic energyﻗﺒﻞ اﻟﺘﺼﺎدم ﻣﺤﺎﻓﻈﺔ conserved أي أن آﻤﻴﺔ اﻟﺤﺮآﺔ ﻗﺒﻞ اﻟﺘﺼﺎدم = آﻤﻴﺔ اﻟﺤﺮآﺔ ﺑﻌﺪ اﻟﺘﺼﺎدم اﻟﻄﺎﻗﺔ اﻟﺤﺮآﻴﺔ ﻗﺒﻞ اﻟﺘﺼﺎدم = اﻟﻄﺎﻗﺔ اﻟﺤﺮآﻴﺔ ﺑﻌﺪ اﻟﺘﺼﺎدم اﻓﺘﺮض ﺟﺴﻤﺎن ﻳﺘﺤﺮآﺎن ﺑﺴﺮﻋﺔ v1iو v2iﻓﻲ اﺗﺠﺎﻩ xاﻟﻤﻮﺟﺐ وﺑﻌﺪ اﻟﺘﺼﺎدم ﺗﻜﻮن ﺳﺮﻋﺘﻬﻤﺎ v1fو v2f ﻋﻠﻰ اﻟﺘﻮاﻟﻲ ،آﻤﺎ هﻮ ﻣﺒﻴﻦ ﻓﻲ اﻟﺸﻜﻞ: y
v2f
m2
m1 v2f
m2
v2i
v1i
m1 x
ﻗﺒﻞ اﻟﺘﺼﺎدم
ﺑﻌﺪ اﻟﺘﺼﺎدم
ﻣﻊ ﻣﻼﺣﻈﺔ أن آﻞ ﺳﺮﻋﺔ ﻣﻮﺟﺒﺔ ) (v> 0ﺗﻌﻨﻲ أن اﻟﺠﺴﻴﻢ ﻳﺘﺤﺮك ﺑﺎﺗﺠﺎﻩ ﻣﺤﻮر xاﻟﻤﻮﺟﺐ وآﻞ ﺳﺮﻋﺔ ﺳﺎﻟﺒﻪ ) (v< 0ﺗﻌﻨﻲ أن اﻟﺠﺴﻴﻢ ﻳﺘﺤﺮك ﺑﺎﺗﺠﺎﻩ ﻣﺤﻮر xاﻟﺴﺎﻟﺐ أو اﻟﻰ اﻟﻴﺴﺎر. ﻣﻦ ﻗﺎﻧﻮن ﺣﻔﻆ آﻤﻴﺔ اﻟﺤﺮآﺔ :
m 1v 1i + m 2v 2 i = m 1v 1f + m 2v 2f
). . . . . . . . . . . . . . . . . . . . . . (1
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Dr. Abdallah M. Azzeer
وﺑﺎﻋﺘﺒﺎر أن اﻟﺘﺼﺎدم ﻣﺮن elastic collisionﻓﺎن اﻟﻄﺎﻗﺔ اﻟﺤﺮآﻴﺔ ﺗﻜﻮن ﻣﺤﺎﻓﻈﻪ ،وﻋﻠﻴﻪ ﻓﺎن : ). . . . . . . . . . . . . . . . . . (2
2 2f
1 m 2v 2
+
2 1f
1 m 1v 2
=
2 2i
1 m 2v 2
+
2 1i
1 m 1v 2
إذا آﺎﻧﺖ v1iو v2iو m1و m2ﻣﻌﻠﻮﻣﺔ ﻣﺴﺒﻘﺎ ,ﻓﺎﻧﻪ ﻳﻤﻜﻦ إﻳﺠﺎد v1fو v2fﻋﻠﻰ اﻟﻨﺤﻮ اﻟﺘﺎﻟﻲ: ﻳﻤﻜﻦ آﺘﺎﺑﺔ اﻟﻤﻌﺎدﻟﺔ ) (1ﻋﻠﻰ اﻟﺼﻮرة :
)) . . . . . . . . . . . . . . . . . . . . . . . .(3
2i
- v
2f
) = m 2 (v
1f
- v
1i
m 1 (v
وآﺬﻟﻚ اﻟﻤﻌﺎدﻟﺔ ) (2ﻋﻠﻰ اﻟﺼﻮرة : 2 2i
) -v أو
2 2f
m 1 (v - v ) = m 2 (v 2 1f
2 1i
)m 1 (v 1i - v 1f )(v 1i + v 1f ) = m 2 (v 2f - v 2i )(v 2f + v 2i ) . . . . . . . . . . . . . . . . . . . . . . . .(4
ﺑﻘﺴﻤﺔ ) (4ﻋﻠﻰ ) (3ﻣﻊ اﻓﺘﺮاض أن v 1i ≠ v 1fو v 2i ≠ v 2f
ﻧﺤﺼﻞ ﻋﻠﻰ :
)v 1i + v 1f = v 2f + v 2i . . . . . . . . . . . . . . . . . . (5 أو
v 1i − v 2i = v 2f − v 1f 28
Dr. Abdallah M. Azzeer
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106 PHYS - CH7
ﻳﻼﺣﻆ ﻣﻦ اﻟﻌﻼﻗﺔ اﻟﺴﺎﺑﻘﺔ أن اﻟﺴﺮﻋﺔ اﻟﻨﺴﺒﻴﺔ ﻟﻠﺠﺴﻤﻴﻦ ﻗﺒﻞ اﻟﺘﺼﺎدم هﻲ ﻧﻔﺴﻬﺎ ﺑﻌﺪ اﻟﺘﺼﺎدم. ﻣﻦ اﻟﻌﻼﻗﺔ ) (5ﻧﺠﺪ أن :
)v 2f = v 1i − v 2i + v 1f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (6 وﺑﺎﻟﺘﻌﻮﻳﺾ ﻋﻦ v 2fﻣﻦ ) (6ﻓﻰ اﻟﻤﻌﺎدﻟﺔ ) (1ﻧﺠﺪ أن :
m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 1i − m 2 v 2i − m 2 v 1f (m 1 − m 2 )v 1i + 2 m 2 v 2i = (m 1 + m 2 ) v 1f ⎞ ⎛ 2m 2 ⎞ ⎛ m − m2 )⎟⎟ v 2i . . . . . . . . . . . . . . (7 ⎜⎜ ⎟⎟ v 1i + v 1f = ⎜⎜ 1 m m + ⎠ 2 ⎠ ⎝ m1+ m2 ⎝ 1
∴
وﺑﺎﻟﺘﻌﻮﻳﺾ ﻋﻦ v 1fﻣﻦ ) (7ﻓﻰ اﻟﻤﻌﺎدﻟﺔ ) (6ﻧﺠﺪ أن ) ﺑﻌﺪ اﻟﺘﺒﺴﻴﻂ اﻟﺮﻳﺎﺿﻲ( :
⎞ ⎛ 2m 1 ⎞ ⎛ m − m1 ⎟⎟ v 1i + ⎜⎜ 2 )⎟⎟ v 2i . . . . . . . . . . . . . . (8 ⎜⎜ = v 2f ⎠ ⎝ m1+ m2 ⎠ ⎝ m1+ m2
29
∴
Dr. Abdallah M. Azzeer
2i
2i
⎞ ⎛ 2m 2 ⎜+ ⎟ v ⎠ ⎝ m 1+ m 2 ⎞⎛m − m1 +⎜ 2 ⎟ v ⎠ ⎝ m 1+ m 2
1i
1i
⎞⎛m − m 2 = ⎜ 1 ⎟ v ⎠ ⎝ m 1+ m 2 ⎞ ⎛ 2m 1 ⎜ = ⎟ v ⎠ ⎝ m 1+ m 2
1f
2f
v
v
ﺣﺎﻻت ﺧﺎﺻﺔ:
) (1إذا آﺎﻧﺖ اﻟﻜﺘﻠﺘﻴﻦ اﻟﻤﺘﺼﺎدﻣﺔ ﻣﺘﺴﺎوﻳﺘﻴﻦ ) ( m1=m2=mﻣﺜﻞ ﺗﺼﺎدم آﺮﺗﻲ ﺑﻠﻴﺎرد ،ﻓﺎن: 2i
= v
1f
v
1i
= v
2f
v
أي أن اﻟﺠﺴﻤﺎن اﻟﻤﺘﺼﺎدﻣﺎن ﻳﺘﺒﺎدﻻن اﻟﺴﺮﻋﺔ .وإذا آﺎن اﻟﺠﺴﻢ اﻟﺜﺎﻧﻲ ﺳﺎآﻨﺎ ﻗﺒﻞ اﻟﺘﺼﺎدم ) (v2i=0ﻓﺎن:
1i
30
= 0
1f
v
= v
2f
v
Dr. Abdallah M. Azzeer
Page 15 15
د.ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ
106 PHYS - CH7
) (2إذا آﺎﻧﺖ ) m1>> m2ﻣﺜﻞ أن ﻳﻜﻮن اﻟﺠﺴﻢ اﻷول آﺮة ﺑﻮﻟﻨﺞ واﻟﺠﺴﻢ اﻟﺜﺎﻧﻲ آﺮة ﺗﻨﺲ ﻃﺎوﻟﺔ ( ،واﻟﺠﺴﻢ اﻟﺜﺎﻧﻲ ﺳﺎآﻨﺎ ﻗﺒﻞ اﻟﺘﺼﺎدم ) (v2=0ﻓﻔﻲ هﺬﻩ اﻟﺤﺎﻟﺔ ﺗﻜﻮن :
≈ v
1f
≈ 2v
2f
1i 1i
v v
أي أن ﺳﺮﻋﺔ آﺮة ﺗﻨﺲ اﻟﻄﺎوﻟﺔ ﺑﻌﺪ اﻟﺘﺼﺎدم ﺗﻜﻮن ﺿﻌﻒ ﺳﺮﻋﺔ آﺮة اﻟﺒﻮﻟﻨﺞ ﻗﺒﻞ اﻟﺘﺼﺎدم ﺗﻘﺮﻳﺒﺎ ،وﺳﺮﻋﺔ آﺮة اﻟﺒﻮﻟﻨﺞ ﺑﻌﺪ اﻟﺘﺼﺎدم ﺗﻘﺮﻳﺒﺎ ﻻ ﺗﺘﻐﻴﺮ ﻋﻤﺎ هﻲ ﻋﻠﻴﻪ ﻗﺒﻞ اﻟﺘﺼﺎدم.
) (3إذا آﺎﻧﺖ ) m1
