Momentum, impulse and collisions | Bài tập môn Vật lý đại cương 1 CTTT | Trường Đại học Bách Khoa Hà Nội

02/07/2021 Linear momentum Linear momentum of a particle:
The time rate of change of the momentum of a particle is equal to the net force acting on the particle
Chapter 8: MOMENTUM, IMPULSE, AND COLLISIONS System of particles
Exercises: 5, 9, 11, 13, 17, 19, 23, 25, 27, 33, 35, 39, 43, 45, 47, 49, 53,57
Problems: 59, 65, 69, 71, 73, 75, 77, 81, 83, 91, 95, 97, 99, 101, 105 Conservation:
Isolated system is a system on which no external force exerts on or the net external
force exerted on it equals to zero.     when   F  0 p  m v m v m v const A A  B B  C C  ...        F  0   F  0 p  m v m v m v const x A Ax  B B x  C Cx  ...  x One Love. One Future. 1 One Love. One Future. 2 1 2 Collision and impulse Center of mass
The of the center of mass of a system of particles: position vector
m x  m x  m x  ... m x m x i i i i
Collision: isolated event in which two or more bodies exert relatively strong forces on each other for a relatively short time. 1 1 2 2 3 3 x         cm m  m  m  ... m M  
m r  m r  m r  ... m r m r 1 2 3 i i i i i Momentum conservation 1 1 2 2 3 3 P  const r    cm elastic collision: x m  m  m  ... m M
m y  m y  m y  ... m y m y 1 2 3 i 1 1 2 2 3 3 i i i i y    cm Energy conservation M  m m  m  m  ... m M 1 2 3 i  i
inelastic collision: momentum conservation P  const  x    
The total momentum of a system:
P  m v  m v  m v  ...  Mv 1 1 2 2 3 3 cm    
Impulse: Measures the strength and duration of the collision force  m v  m v  m v  ... P
The velocity of the center of mass of the system of particles 1 1 2 2 3 3 v   t cm 2      M M
F  const  J  Fdt  F(t  t )  Ft   2 1
If the net external force on the system is not zero, the acceleration of the center of mass is: 2 t     F  M a t ext cm J  Fdt 1 t2    
Systems with varying mass (rocket+exhaust products) 1 t F  F(t)  J  F(t)dt  1 t Impulse-momentum theorem t  2 t2 t2 t2    dv      J  Fdt  madt  m
dt  mdv  mv  mv  p  p     2 1 2 1 dt dM < 0 1 t 1 t 1 t 1 t
The change in a particle’s momentum during a time interval equals the impulse of the net force that acted on the particle during that interval. Before After dt One Love. One Future. 3 One Love. One Future. 4 3 4 02/07/2021
8.59 The net force acting on a 2.00-kg discus while it is being thrown is t2.𝚤⃑+(+t) 𝚥⃑, where  = 25.0 N/s2,  = 30.0 N and
 = 5.0 N/s. If the discus was originally at rest, what is its velocity after the net force has acted for 0.500 s? Express your
answer in terms of the unit vectors 𝚤⃑ and 𝚥⃑  p   Fdt  v p  mv                 2       3 2 p Fdt t i (
t) j dt   t  i   t  t  j  3   2  p            3 2 v    t  i   t  t  j  0.52i  7.82 j m  3m   2m m  One Love. One Future. 7/2/2021 Dang Duc Vuong - SEP - HUST 5 One Love. One Future. 6 5 6
8.65(70). A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases,
a) Let A be the 25.0 kg mass and B be the car (mass 175 kg)
the car initially has a total mass (car and contents) of 200 kg and is traveling east with a velocity of magnitude 5.00 m/s.
Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks. (a) A 25.0-kg mass is
thrown sideways out of the car with a velocity of magnitude 2.00 m/s relative to the car's initial velocity. (b) A 25.0-kg mass
is thrown backward out of the car with a velocity of 5.00 m/s relative to the initial motion of the car. (c) A 25.0-kg mass is
thrown into the car with a velocity of 6.00 m/s relative to the ground and opposite in direction to the initial velocity of the car.
(b) A 25.0-kg mass is thrown backward out of the car z z z y y y v v v x x x (c) (a) (b) (c) Identify:
Use a coordinate system attached to the ground. Take the x-axis to be east (along the tracks) and the y-axis to be north
(parallel to the ground and perpendicular to the tracks). Then Px is conserved and Py is not conserved, due to the sideways
force exerted by the tracks, the force that keeps the handcar on the tracks. One Love. One Future. 7 One Love. One Future. 8 7 8 02/07/2021
8.69(74). A 0.150-kg frame, when suspended from a coil spring, stretches the spring 0.050 m. A 0.200-kg
8.71(76). A Ricocheting Bullet. 0. 100-kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 g, traveling
lump of putty is dropped from rest onto the frame from a height of 30.0 cm (Fig.). Find the maximum
horizontally at 350 m/s, strikes the stone and rebounds horizontally at right angles to its original direction with a speed of
distance the frame moves downward from its initial position. the spring to find the force constant k of the
250 m/s. (a) Compute the magnitude and direction of the velocity of the stone after it is struck. (b) Is the collision
spring. Let s be the amount the spring is stretched. perfectly elastic? Energy is conserved d K  U  W  K  U 0 d 1 1 other 2 2 1 d M  m 1 d1 2 K  v 1 2 2 d2 s
U  M  mgd  1 2  k(d  d ) 1 1 1 0 2 v0=0 m W  0 h M P other d x is conserved K  0 m v  m v  m v  m v A A1x B B1x A A 2 x B B 2 x 2 1 m v
U  M  m g d  k(d  d ) A A1x  m v  m v  v   21.0 m / s A A1x B B2 x B 2 x   2    2  2 k(d  d )  Mg v v 2 0 1 2 2 m 1 0 B P M  m 2 2 M 0.150 y is conserved 1 P    2 2 v v v 25.8m / s 2 y is conserved 
v  M  m g(d  d )  k  d  d  d  d   0 B B1x B 2 y 2   2 1  1 0   2 0 k  g  9.81  29.4N / m mv  M.0  M  m v 2 2   m v  m v  m v  m v A A1y B B1y A A 2 y B B 2 y v 1   s 0.050 2 B 2 y o tan    0.7143;   35.5 mv M  m 1 m        A 0  m v  m v  v  v  15.0 m / s v A A 2 B B 2 y B 2 y A 2   2 2 v  v  2gh  v  2gh v M m gd k s d s  0 2    2 2 2 1 v    B 2 y 1 2 2 2 m 0 1 M  m B
 v  2(9.81)(0.300)  2.425 m / s 2
14.7d 1.96d  0.3362  0        not elastic 1   v 1.386 m / s
if the collision perfectly elastic  Energy is conservations? K K K 147J 0 2 1 2   d  0.0667m One Love. One Future. 9 One Love. One Future. 10 9 10
8.(73)78. Two identical masses are released from rest in a smooth hemispherical bowl of radius R,
8.75(80). A 20.00-kg lead sphere is hanging from a hook by a thin wire 3.50 m long, and is free to swing in a complete circle.
from the positions shown in Fig. You can ignore friction between the masses and the surface of the
Suddenly it is struck horizontally by a 5.00-kg steel dart that embeds itself in the lead sphere. What must be the minimum
bowl. If they stick together when they collide, how high above the bottom of the bowl will the masses
initial speed of the dart so that the combination makes a complete circular loop after the collision? go after colliding?
The steel dart collides the lead sphere: non elastic Conservation of momentum
IDENTIFY: Apply conservation of energy to the motion before and after the collision and apply conservation of m 5 v momentum to the collision. mv  Mv  m  M v A1  v  v  v  A1 A1   1 1 m  M A1 A1 20  5 5 A
The combination makes a complete circular loop after the collision2 v
at the top of the circular loop, the object has speed v 2 a  2 rad R R
T  (m  M)g  (m  M)a 3 v A B rad 2 g   v  gR h v A+B 2 v T  0 R 1 2
use conservation of energy with point 1 at the bottom of the loop and point 2 A body: Conservation of energy Conservation of momentum Conservation of energy at the top of the loop 1 1 1 M 2 m v  m gR  v  2gR applied to the collision 1 2 2 2 m v  0  m v  m g 2R AB 1 AB 2 AB   A 1 A 1 2 m  m v  m  m gh m A B  2  A B  2 2 m v  0  m  m v 2 2 A 1  A B  2 v R m  m  M  v  65.5 m / s 1A   m v 2 h   AB v gR 2 A 1 v  v  2g 4 v 2 m  m  1 2 A1 v  A B 1 5 One Love. One Future. 11 One Love. One Future. 12 11 12 02/07/2021
8.77(83). A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass
(b) What is the decrease in kinetic energy of the bullet?
0.800 kg, initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 120 m/s.
The block slides a distance of 45.0 cm along the surface from its initial position. (a) What is the coefficient of kinetic friction
between block and surface? (b) What is the decrease in kinetic energy of the bullet? (c) What is the kinetic energy of the block
(c) What is the kinetic energy of the block at the instant after the bullet passes through it
at the instant after the bullet passes through it?
Bullet is fired into a wooden block Momentum is conserved
Motion of the block after the collision. One Love. One Future. 13 One Love. One Future. 14 13 14
8.81(87). In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left
at a speed of 5.00 m/s (Fig.). You can ignore friction between the cart and the floor. A 15.0-kg
package slides down a chute that is inclined at 37° from the horizontal and leaves the end of the
chute with a speed of 3.00 m/s. The package lands in the cart and they roll off together. If the lower
end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what are (a) the speed
of the package just before it lands in the cart and (b) the final speed of the cart?
Apply conservation of energy to the motion of the package    1 1 Mv  m v  m v 2 2 2
mv  mgh  mv  v  v  2gh  9.35 m / s cm A A B B      0 1 1 0   2 2 2 v 2 2
v  v  v  v  v  v  2v v v A A cm A A cm A cm 0
In the collision between the package and the cart momentum is conserved in the v cm      A v 2 2 2
v  v  v  v  v  v  2v v y B B cm B B cm B cm horizontal direction. B 1 1 1   1   Center mass 2 2 K  m v  m v  m  2 2
v  v  2v v   m  2 2 v  v  2v v A A B B A A cm A cm B B cm B cm 
P  const  mv  MV  M  m V m 2 2 2 2 x 1x 0   1 A m x B 1 1 v  v  v cos  K      2 2
m v  m v   m  m  2 v  m v  m v v A A B B A B cm  A A B B  1x 0 x 0 v cm 2 2 1  mv  MV mv cos  MV      1 1 1x 0 0 0  V    3.29 m / s
m v  m v v  Mv .v  0  K  m v  m v  Mv A A B B  cm cm cm  2 2 A A B B  2 V 1 M  m M  m   cm 2 2 0 1 The asteroids col ide 2 K  Mv  const KE of internal motion v min 2 cm KE of center motion 1 One Love. One Future. 15 One Love. One Future. 16 15 16 02/07/2021
8.91(97). Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is
8.95(101). You are standing on a concrete slab that in turn is resting on a frozen lake. Assume there is no friction between the
initially traveling at 15.0 m/s and is deflected 25.0o from its initial direction. Assume that the collision is perfectly elastic.
slab and the ice. The slab has a weight five times your weight. If you begin walking forward at 2.00 m/s relative to the ice,
Find the final speed of each puck and the direction of B's velocity after the collision. [Hint: Use the relationship derived in
with what speed, relative to the ice, does the slab move? part (d) of Problem 8.96.]
The is no friction between slab and ice;
The system of slab and man is an isolated system;   ? 2.00 m/s
Use the momentum conservation with the motion of man on slab m  m  m 25o   1 2  m 1
P  const  0  mv  Mv  v   v    2 m / s 1 2 2 1   M 5
The slab moves backward with speed of 0.4 m/s
Ox : v  v cos  v cos   1 A A2 B2 P  const   m v   m v   m v A 1 A A A2 B B2
Oy : 0  v sin  v cos  A2 B 2 Perfectly elastic collision 1 v  13.7 m / s A2   2 1 2 1 2 K  const  m v  m v  m v   ;   65o A 1 A A2 A2 B B 2 2 2 2 v  6.4 m /  s B 2   One Love. One Future. 17 One Love. One Future. 18 17 18
8.97(103). A fireworks rocket is fired vertically upward. At its maximum height of 80.0 m, it explodes and breaks into two
8.99(105). A Nuclear Reaction. Fission, the process that supplies energy in
pieces, one with mass 1.40 kg and the other with mass 0.28 kg. In the explosion, 860 J of chemical energy is converted to
nuclear power plants, occurs when a heavy nucleus is split into two medium-
kinetic energy of the two fragments. (a) What is the speed of each fragment just after the explosion? (b) It is observed
sized nuclei. One such reaction occurs when a neutron colliding with a 235 U
that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they
(uranium) nucleus splits that nucleus into a
land? Assume that the ground is level and air resistance can be ignored.
141Ba (barium) nucleus and a 92Kr
(krypton) nucleus. In this reaction, two neutrons also are split off from the m  1.40kg m  0.28kg 1 2
At the highest point, the velocity of the rocket is vi=0 original     
235U. Before the collision, the arrangement is as shown in Fig.a. After  
* The momentum conservation: p  p  0  m v  m v the collision, the i f 1 1 2 2 v v 1 2
141Ba nucleus is moving in the +z-direction and the 92Kr
 Two parts move in opposite directions  m v  m v  0
nucleus in the -z-direction. The three neutrons are moving in the xy-plane, as 1 1 2 2 1 1
shown in Fig.b. If the incoming neutron has an initial velocity of magnitude ** Energy conservation: 2 2
KE  m v  m v  860 J 1 1 2 2   h 2 2
3.0x103 m/s and a final velocity of magnitude 2.0x103 m/s in the directions v 14.3 m / s
shown, what are the speeds of the other two neutrons, and what can you say 1    v  71.5 m/ s 
about the speeds of the 141Ba and 92Kr nuclei? (The mass of the 141Ba nucleus is 2  
approximately 2.3x10-25 kg, and the mass of
(b) Two parts land at the same time  v 92Kr is about 1.5x10-25 kg.)
1, v2 are in horizontal direction 2h Time to hit the ground: t 
 4.04s  AB  v  v t  347 m 1 2    A B g One Love. One Future. 19 One Love. One Future. 20 19 20 02/07/2021
Consider the process in Oxyz coordinate system (Figure). Note that positive z-
direction is from back to front of the figure.
8.101(107). The coefficient of restitution  for a head-on collision is defined as the ratio of the relative speed after the collision
Use the momentum conservation for the collision we have:
to the relative speed before. (a) What is  for a completely inelastic collision? (b) What is  for an elastic collision? (c) A   p  p
ball is dropped from a height h onto a stationary surface and rebounds back to a height H i f 1. Show that Є = 𝐻1/ℎ (d) A        m v  m v  m v  m v  M v  M v 
properly inflated basketball should have a coefficient of restitution of 0.85. When dropped from a height of 1.2 m above a n oni n on f n en 1  n en2 Ba Ba Kr Kr ven 1
solid wood floor, to what height should a properly inflated basketball bounce? (e) The height of the first bounce is H Project on Oxyz we have: 1. If 
is constant, show that the height of the n-th bounce is H On Ox: m v  m v cos10  m v cos 45  m v cos30
n = 2nh. (f) If  is constant, what is the height of the eighth bounce n oni n on f n en 1  n en2   v v
of a properly inflated basketball dropped from 1.2 m? oni on f  v  v cos10  v cos 45  v cos30 oni on f en 1  en2
(a) completely inelastic collision  = 0 On Oy: 0  m v sin10  m v sin 45  m v sin 30 n on f n en 1  n en2      0 v sin10 v sin 45 v sin 30 (b) elastic collision  = 1 on f en 1  en2 ven2 (c)  v On Oz: 0  M v  M v
Speed of ball just before collision: v 2gh i f H Ba Ba Kr Kr v  221 m / s 1     en 1    v  2gH (a) v 
The velocity just after collision: f 1 v h
on-i= 3.0x103 m/s; von-f= 2.0x103 m/s ;  3 i v  1.0110 m / s  en2   (d) We have: ε=0.85 2  H   h  0.87m 1
(e) H   h  H   H   h  H  n  h  n  h 1 2 1  2 2 2 2 n  2 2 n + U → 3n + Ba + Kr v M (b) We have: 0  M v  M v Kr Ba    1.5 Coefficient of restitution  Ba Ba Kr Kr v M (f) n = 8
 H   h  1.20.8516 2 8  0.089m 8 Є=v Ba Kr f/vi; One Love. One Future. 21 One Love. One Future. 22 21 22
*8.105(111). A Multistage Rocket. Suppose the first stage of a two-stage rocket has total mass 12,000 kg, of which 9000 kg is
a) The total initial mass of the rocket is Mi = 12,000 kg+1000 kg = 13,000 kg
fuel. The total mass of the second stage is 1000 kg, of which 700 kg is fuel. Assume that the relative speed vex of ejected
The fuel: 9000 kg+700 kg = 9700 kg
material is constant, and ignore any effect of gravity. (The effect of gravity is small during the firing period if the rate of fuel
consumption is large). (a) Suppose the entire fuel supply carried by the two-stage rocket is utilized in a single-stage rocket
The mass Mf left after all the fuel is burned is = 13,000 kg- 9700 kg= 3300 kg M M 13000
with the same total mass of 13,000 kg. In terms of v i i
ex, what is the speed of the rocket, starting from rest, when its fuel is v  v  v ln  v  v ln  v ln 1.37v f i ex f ex ex ex
exhausted? (b) For the two-stage rocket, what is the speed when the fuel of the first stage is exhausted if the first stage M M 3300 f f
carries the second stage with it to this point? This speed then becomes the initial speed of the second stage. At this point, the
b) vi = 0; Mi = 13 000 kg; Mf1 = Mi – Mfuel1 = 4000 kg;
 The final speed of the 1st stage:
second stage separates from the first stage. (c) What is the final speed of the second stage? (d) What value of vex is required M 13000 i v  v ln  v  v ln  0  1.18v
to give the second stage of the rocket a speed of 7.00 km/s? f1 ex i ex ex M 4000 f (c) We have: v For the single-stage rocket
i2 = vf1 = 1.18vex; Mi2 = Mstage2=1000 kg; Mf2 = Mi2-Mfuel2 = 300 kg;
Note that the 1st stage separated (not exhausted) with speed of 0 m/s relative to the 2nd stage, so it does not change the Mi v  v  v ln speed of stage 2. f i ex M M 1000 f
 The final speed of the 2nd stage: i2 v  v ln  v  v l 1.18v  2.38v M f 2 e i2 ex ex ex i: the total initial mass M 3300
Rocket propulsion. (a) The initial mass of the rocket plus all its fuel is M f 2
Mf: the mass left after all the fuel is burned
+Δm at a time t, and its speed is v. (b) At a time t+Δt, the rocket’s mass has vf2
been reduced to M and an amount of fuel Δm has been ejected. The v
(d) To have final speed of the 2nd stage is 7 km/s=7000 m/s  v   2940 m / s  2.94 km / s ex     ex: the ejected material
rocket’s speed increases by an amount Δv. 2.38 One Love. One Future. 23 One Love. One Future. 24 23 24 02/07/2021 www.hust.edu.vn
Thank you for your attentions!25 25