Dynamics : 1 Dynamics Section 5 – Conservation of Momentum
Reference: Chpt 6 Momentum and Collisions pg 154 -183 College Physics - Serway & Faughn
Content : 2 Content 5. Law of Conservation of momentum
5.1 Law of Conservation of momentum
5.2 Elastic Collision
5.3 Inelastic Collision
Learning Outcomes - 2 : 3 Learning Outcomes - 2 state the principle of conservation of momentum.
apply the principle of conservation of momentum to solve simple problems including elastic and inelastic interactions between two bodies in one dimension.
recognize that, for a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation.
show an understanding that, whilst momentum of a system is always conserved in interactions between bodies, some change in kinetic energy usually takes place.
5.1 Principle of Conservation of Momentum : 4 5.1 Principle of Conservation of Momentum Consider mass A traveling with velocity u1 collides with mass B with velocity u2. During the collision, the two masses are in contact and exert a force F on each other. The time of impact is ?t.
Slide 5 : 5 5.1 Principle of Conservation of Momentum Consider mass A traveling with velocity u1 collides with mass B with velocity u2. During the collision, the two masses are in contact and exert a force F on each other. The time of impact is ?t.
5.1 Principle of Conservation of Momentum : 6 5.1 Principle of Conservation of Momentum m1 u1 m2 u2 -F2 F1 m2 v2 m1 v1 By N2L: For A: F2 = ?p1 /?t For B: F1 = ?p2 /?t By N3L: F1 = - F2 = (m1v1 – m1u1)/?t = (m2v2 – m2u2)/?t = -(m1v1 – m1u1)/?t (m2v2 – m2u2)/?t
5.1 Principle of Conservation of Momentum : 7 5.1 Principle of Conservation of Momentum m1 u1 m2 u2 -F2 F1 m2 v2 m1 v1 = -(m1v1 – m1u1) (m2v2 – m2u2) Rearranging we get m1u1 + m2u2 = m2v2 + m1v1 Total initial momentum = total final momentum
5.1 Principle of Conservation of Momentum : 8 5.1 Principle of Conservation of Momentum The Principle of Conservation of Momentum states that the provided no external force acts on the system. total momentum of the
system before the collision is equal to the total momentum of the system after the collision,
Simple Collisions in 1-Dimension : 9 Simple Collisions in 1-Dimension Example 5.1.1
A ball A of mass 0.10 kg moving with a velocity of 6.0 m s-1 collides directly with ball B of mass 0.20 kg initially at rest. Calculate their common velocity if both balls move off together. 0.10 kg 0.20 kg 0.30 kg Before collision: After collision:
Example 5.1.1 : 10 Example 5.1.1 0.10 kg UA=
6.0 m s-1 0.20 kg V 0.30 kg Before collision: After collision: Ans: Taking ? as positive direction.
Since there is no ext force acting on the collision, by Principle of Conservation of Momentum mAuA + mBuB = (mA + mB)V
0.10x6.0 = 0.30 V
? V = 2.0 m s-1
Example 5.1.2 : 11 Example 5.1.2 A 7.0 kg bowling ball of velocity u collides head-on with a 2.0 kg bowling pin. The pin flies forward with a velocity of 3.0 m s-1 and the ball continues with a velocity of 1.8 m s-1. What was the initial velocity u of the ball? 7.0 kg 2.0 kg Before collision: After collision:
Example 5.1.2 : 12 Example 5.1.2 7.0 kg u 2.0 kg 1.8 m s-1 3.0 m s-1 Ans: Taking ? as +ve direction.
By P of C of Momentum m1u + m2u2 = m1v1 + m2v2
7.0u = 7.0x1.8 + 2.0x3.0
? u = 2.7 m s-1 Before collision: After collision:
Example 5.1.3 : 13 Example 5.1.3 A cue ball of mass 0.15 kg initial velocity 5.0 m s-1 collides with a billiard ball of mass 0.20 kg initially at rest. After the collision, the billiard ball moves with a velocity of 4.0 m s-1. (a) Calculate the velocity of the cue ball after the collision.
(b) Calculate the total k.e. before and after the collisions. Before collision: After collision:
Example 5.1.3 : 14 Example 5.1.3 Before collision: After collision: m1u1 + m2u2 = m1v1 + m2v2
0.15x5.0 = 0.15 x v1 + 0.20x4.0
? v1 = - 0.33 m s-1 The negative sign indicates that the cue ball is now travelling to the left. Ans: Taking ? as positive direction.
By P of C of Momentum
Example 5.1.3 : 15 Example 5.1.3 Before collision: After collision: Tot KE aft coll = ½ m1v12 + ½ m2v22
= ½ x0.15x0.332 + ½ x0.20x4.02
= 1.6 J Hence, total KE is lost during the collision.
Such a collision is known as an inelastic collision. (b) Tot KE b4 coll = ½ m1u12
= ½ x0.15x5.02 = 1.9 J
Example 5.1.4 : 16 Example 5.1.4 A 30.0 g bullet is fired vertically at 200 m s-1 into a block of wood of mass 0.150 kg initially at rest. How high does the combination rise after the collision, assuming that the bullet embeds itself in the wood? 0.150 kg Before collision After collision
Example 5.1.4 : 17 Example 5.1.4 0.150 kg u1=
200 ms-1 V Before collision After collision 0.0300 kg Ans: By P of C of Momentum: m1u1 + m2u2 = (m1 + m2)V
0.0300x200 = (0.0300 + 0.150)V
? V = 33.3 m s-1
Example 5.1.4 : 18 Example 5.1.4 V Vf = 0 Using v2 = u2 + 2as
0 = V2 + 2(-g)s
0 = 33.32 + 2(-9.81)s
? s = 56.5 m
Example 5.1.5: Finding speed of bullet using a ballistic pendulum : 19 Example 5.1.5: Finding speed of bullet using a ballistic pendulum A 7.0 g bullet is fired into a 1.5 kg ballistic pendulum. The bullet is embedded in the block and it rises to a maximum height of 12 cm. Find the initial speed of the bullet. Before collision After collision
Example 5.1.5: Finding speed of bullet using a ballistic pendulum : 20 Example 5.1.5: Finding speed of bullet using a ballistic pendulum Before collision u 0.0070kg V After collision m1u1 + m2u2 = (m1 + m2)V
m1u = (m1 + m2)V - - -(1) By P of C of Momentum
Example 5.1.5: Finding speed of bullet using a ballistic pendulum : 21 Example 5.1.5: Finding speed of bullet using a ballistic pendulum V h After collision For block to rise to height h,
KE of block is changed into
GPE at height h. KE = GPE
½ (m1+m2)V2 = (m1+m2)gh
V2 = 2gh - - - (2)
Example 5.1.5: Finding speed of bullet using a ballistic pendulum : 22 Example 5.1.5: Finding speed of bullet using a ballistic pendulum m1u = (m1 + m2)V - - -(1) V2 = 2gh - - - (2) u = ((m1 + m2)/m1)(?2gh)
= ((0.0070+1.5)/0.0070)(?2x9.81x0.12)
= 330 m s-1 Hence by measuring the height reached by the pendulum, we can deduced the speed u of the bullet. In this expt, u = 330 m s-1
Example 5.1.6: Recoil of a gun : 23 Example 5.1.6: Recoil of a gun A gun of mass 1.0 kg has a bullet of mass 0.1 kg inside. The bullet leaves the gun when fired at a velocity of 200 m s-1. Calculate the velocity of the gun after firing. Before: After:
Example 5.1.6: Recoil of a gun : 24 Example 5.1.6: Recoil of a gun Before: U = 0 -vgun vb After: Mgun mb By P of C of Momentum (Mgun + mb)U = mbvb - MgunVgun 0 = mbvb - MgunVgun ? MgunVgun = mbvb - - - (1)
Example 5.1.6: Recoil of a gun : 25 Example 5.1.6: Recoil of a gun MgunVgun = mbvb
Vgun = 0.1x200/1.0
= 20 m s-1 Note:
From Eqn (1): From this eqn, we can deduce that in an explosive situation, the big mass will have a smaller velocity, while the small mass will have larger velocity.
Example 5.1.7: Radioactive Decay by ?-particle : 26 Example 5.1.7: Radioactive Decay by ?-particle Consider a radioactive decay of
The parent nucleus U is initially at rest. The ?-particle is found to be emitted at a speed of 5.0 x 106 m s-1. Calculate the speed of the daughter nucleus Th. Before: After:
Example 5.1.7: Radioactive Decay by ?-particle : 27 Example 5.1.7: Radioactive Decay by ?-particle Before: 234Th -VTh 4He VHe =
5.0 x 106 m s-1 After: By Cons of Momentum:
Total momentum before = total momentum after 0 = mHevHe - mThvTh
Example 5.1.7: Radioactive Decay by ?-particle : 28 Example 5.1.7: Radioactive Decay by ?-particle Note: The big mass will have a smaller velocity, while the small mass will have larger velocity.
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5.2 Elastic and Inelastic Collisions : 30 5.2 Elastic and Inelastic Collisions Collisions Total momentum is conserved.
Total kinetic energy is also conserved. Total momentum is conserved.
Total k.e not conserved.
5.2.1 Elastic Collisions : 31 5.2.1 Elastic Collisions Consider two spheres A and B colliding elastically with each other. The initial velocities of A and B are u1 and u2 respectively. After the collision, the A and B move off with v1 and v2 respectively. Before: After
5.2.1 Elastic Collisions : 32 5.2.1 Elastic Collisions Before: m1
A u1 m2
B u2 v1 v2 After: m1u1 + m2u2 m1
A m2
B ½ m1u12 + ½ m2u22 - - - (1) - - (2)
5.2.1 Elastic Collisions : 33 5.2.1 Elastic Collisions Rewriting eqn (2) gives: - - (2) Rewriting eqn (1) gives: - - - (1) Eqn (2) ? (1) gives: u1 + v1 = v2 + u2
5.2.1 Elastic Collisions : 34 5.2.1 Elastic Collisions Before: m1
A u1 m2
B u2 v1 v2 After: m1
A m2
B For elastic collisions, we can deduce that: The relative
speed of approach = the relative
speed of separation
5.2.1 Elastic Collisions : 35 5.2.1 Elastic Collisions In addition, solving re-expressing equations (1) and (2) we can arrive at
5.2.1 Elastic Collisions : 36 5.2.1 Elastic Collisions Three specific situations: 1. Two identical masses, m1 = m2 collide, eg two billiards collide: v1 = u2 v2 = u1 That means that the two masses exchange velocities after the collision. and
5.2.1 Elastic Collisions : 37 5.2.1 Elastic Collisions 2. A extremely big mass, m1 collides with a very small mass m2 ie. m1 >> m2 , eg. bowling ball hit ping pong v1 ? u1 and v2 ? 2u1 The big mass m1 continues with its initial speed, while the small mass m2 bounces off with twice the speed of m1.
5.2.1 Elastic Collisions : 38 5.2.1 Elastic Collisions 3. A very small mass m1 collides with a very big mass m2 , ie. m2 >> m1, eg. a ping pong striking a wall v1 ? -u1 and v2 ? u2 The small mass m1 rebounds with the same speed,
and the big mass continues to move with its initial speed.
Example 5.2.1: Elastic Collision : 39 Example 5.2.1: Elastic Collision A 25.0 kg ball moving to the right at 20.0 m s-1 overtakes and collides elastically with a 10.0 kg ball moving in the same direction at 15.0 m s-1. Find the velocity of each object after the collision. Before: m1
25 kg m2
10 kg After:
Example 5.2.1: Elastic Collision : 40 Example 5.2.1: Elastic Collision Before: m1
25 kg u1 =
20 ms-1 m2
10 kg u2 =
15 ms-1 v1 v2 m1
25 kg m2
10 kg After: Cons of Momentum: m1u1 + m2u2 = m1v1 + m2v2 - - - (1) The relative
speed of approach = the relative
speed of separation u1 – u2 = v2 – v1 - - - (2)
Example 5.2.1: Elastic Collision : 41 Example 5.2.1: Elastic Collision m1u1 + m2u2 = m1v1 + m2v2 - - - (1) 25x20 + 10x15 = 25v1 + 10v2 - - - (1) u1 – u2 = v2 – v1 - - - (2) 20 – 15 = v2 – v1 - - - (2) Solving (1) and (2), we get: v1 = 17.1 m s-1 and v2 = 22.1 m s-1
Example 5.2.2: Elastic Collision : 42 Example 5.2.2: Elastic Collision A 10.0 kg ball moving to the right at 10.0 m s-1 makes an elastic head-on collision with a 15.0 kg ball moving to the left at 30.0 m s-1. Find the velocity of each object after the collision. Before: After:
Example 5.2.2: Elastic Collision : 43 Example 5.2.2: Elastic Collision Before: m1
10 kg u1 =
10 ms-1 m2
15 kg -u2 =
-30 ms-1 v1 v2 m1
10 kg m2
15 kg After Cons of Momentum: m1u1 - m2u2 = m1v1 + m2v2 - - - (1) The relative
speed of approach = the relative
speed of separation u1 – (-u2) = v2 – v1 - - - (2)
Example 5.2.2: Elastic Collision : 44 Example 5.2.2: Elastic Collision m1u1 - m2u2 = m1v1 + m2v2 - - - (1) u1 – (-u2) = v2 – v1 - - - (2) 10x10 - 15x30 = 10v1 + 15v2 10 – (-30) = v2 – v1 - - - (2) - - - (1) 40 = v2 – v1 Solving gives: v1 = -38.0 m s-1 (moving to the left)
v2 = 2.00 m s-1
Example 5.2.3: Elastic Collision of the same masses : 45 Example 5.2.3: Elastic Collision of the same masses Two billiard balls of identical masses move toward one another and make an elastic collision. If the initial velocities of the balls are + 0.30 m s-1 and -0.20 m s-1, what is the velocity of each ball after the collisions? Before: After:
Example 5.2.3: Elastic Collision of the same masses : 46 Example 5.2.3: Elastic Collision of the same masses Before: m1 u1 =
0.30 ms-1 m2 v1 v2 m1 m2 After: -u2 =
-0.20 ms-1 Cons of Momentum: m1u1 - m2u2 = m1v1 + m2v2 - - - (1) The relative
speed of approach = the relative
speed of separation u1 – (-u2) = v2 – v1 - - - (2)
Example 5.2.3: Elastic Collision of the same masses : 47 Example 5.2.3: Elastic Collision of the same masses m1u1 - m2u2 = m1v1 + m2v2 - - - (1) u1 – u2 = v1 + v2 - - - (1) Since m1 = m2, then eqn becomes u1 – (-u2) = v2 – v1 - - - (2) Solving gives: v1 = -u2
= - 0.20 m s-1 (moving to the left) v2 = u1 = 0.30 m s-1 Note that the two masses exchange speeds.
Slide 48 : 48 If I have seen further it is by standing on the shoulders of giants.
Isaac Newton, Feb 5, 1675
Slide 49 : 49 On the Shoulders of Giants: The Great Works of Physics and Astronomy (Paperback) by Stephen W. Hawking