KINEMATICS OF PARTICLES - Bharati Vidyapeeth's College of...
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KINEMATICS OF PARTICLES
Transcript of KINEMATICS OF PARTICLES - Bharati Vidyapeeth's College of...
KINEMATICS OF PARTICLES
INTRODUCTION
• Kinematics– Study of the geometry of motion. – Relates displacement, velocity, acceleration, and time
without reference to the cause of motion.
• Particles– Not strictly to small particles – possibly as large as
cars, rockets or airplanes.– The entire bodies will analyze, any rotation to the
centre will be neglected
Kinematics of Particles Cases
• Curvilinear motion: position, velocity, and acceleration of a particle as it moves along a curved line in two or three dimensions.
• Rectilinear motion: position, velocity, and acceleration of a particle as it moves along a straight line.
RECTILINEAR MOTION
SUMMARY OF KINEMATIC RELATIONS:RECTILINEAR MOTION
• Differentiate position to get velocity and acceleration.
v = dx/dt ; a = dv/dt or a = v d2x/dt2; a = v dv/dx
• Integrate acceleration for velocity and position.
Note that so and vo represent the initial position and velocity of the particle at t = 0.
Velocity:
=t
o
v
vo
dtadv =s
s
v
v oo
dsadvvor =t
o
s
so
dtvds
Position:
v =
Example
2312 ttdt
dxv ==
tdt
xd
dt
dva 612
2
2
===
- at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0
- at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2
• What are x, v, and a at t = 2 s ?
• Note that vmax occurs when a=0, and that the slope of the velocity curve is zero at this point.
• What are x, v, and a at t = 4 s ?
326 ttx =
Determination of the Motion of a Particle
• We often determine accelerations from the forces applied (kinetics will be covered later)
• Generally have three classes of motion
- acceleration given as a function of time, a = f(t)
- acceleration given as a function of position, a = f(x)
- acceleration given as a function of velocity, a = f(v)
Acceleration as a function of time, position, or velocity
a a t= 0 0
v t
v
dv a t dt= ( )dv
a tdt
=
v dv a x dx=
0 0
v x
v x
vdv a x dx= a a x=
and dx dv
dt av dt
= =
dv
v a vdx
=
0 0
v t
v
dvdt
a v=
0 0
x v
x v
vdvdx
a v=
a a v=
( )dv
a vdt
=
If…. Kinematic relationship Integrate
Example Ball tossed with 10 m/s vertical velocity from
window 20 m above ground.
Determine:
• velocity and elevation above ground at time t,
• highest elevation reached by ball and corresponding time, and
• time when ball will hit the ground and corresponding velocity.
SOLUTION:
• Integrate twice to find v(t) and y(t).
• Solve for t when velocity equals zero (time for maximum elevation) and evaluate corresponding altitude.
• Solve for t when altitude equals zero (time for ground impact) and evaluate corresponding velocity.
tvtvdtdv
adt
dv
ttv
v
81.981.9
sm81.9
00
2
0
==
==
ttv
=
2s
m81.9
s
m10
2
21
00
81.91081.910
81.910
0
ttytydttdy
tvdt
dy
tty
y
==
==
2
2s
m905.4
s
m10m20 ttty
=
SOLUTION:
• Integrate twice to find v(t) and y(t).
• Solve for t at which velocity equals zero and evaluate
corresponding altitude.
0s
m81.9
s
m10
2=
= ttv
s019.1=t
22
2
2
s019.1s
m905.4s019.1
s
m10m20
s
m905.4
s
m10m20
=
=
y
ttty
m1.25=y
• Solve for t at which altitude equals zero and
evaluate corresponding velocity.
0s
m905.4
s
m10m20 2
2=
= ttty
s28.3
smeaningles s243.1
=
=
t
t
s28.3s
m81.9
s
m10s28.3
s
m81.9
s
m10
2
2
=
=
v
ttv
s
m2.22=v
Problem
Brake mechanism used to reduce gunrecoil consists of piston attached tobarrel moving in fixed cylinder filledwith oil. As barrel recoils with initialvelocity v0, piston moves and oil isforced through orifices in piston,causing piston and cylinder todecelerate at rate proportional to theirvelocity. Determine v(t), x(t), and v(x).
kva =
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
• Integrate v(t) = dx/dt to find x(t).
• Integrate a = v dv/dx = -kv to find v(x).
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
000
ln
v t
v
v tdv dva kv k dt kt
dt v v= = = =
ktevtv = 0
• Integrate v(t) = dx/dt to find x(t).
0
0 0
00 0
1
kt
tx t
kt kt
dxv t v e
dt
dx v e dt x t v ek
= =
= =
ktek
vtx = 10
• Integrate a = v dv/dx = -kv to find v(x).
kxvv
dxkdvdxkdvkvdx
dvva
xv
v
=
====
0
00
kxvv = 0
• Alternatively,
=
0
0 1v
tv
k
vtx
kxvv = 0
00 or
v
tveevtv ktkt ==
ktek
vtx = 10with
and
then
Uniform Rectilinear Motion
• During free-fall, aparachutist reachesterminal velocity when herweight equals the dragforce. If motion is in astraight line, this is uniformrectilinear motion.
For a particle in uniform rectilinear motion, the acceleration is zero and the velocity is constant.
vtxx
vtxx
dtvdx
vdt
dx
tx
x
=
=
=
==
0
0
00
constant
Uniformly Accelerated Rectilinear Motion
If forces applied to a bodyare constant (and in aconstant direction), thenyou have uniformlyaccelerated rectilinearmotion.
Another example is free-fall when drag is negligible
UNIFORMLY ACCELERATED RECTILINEAR MOTION
For a particle in uniformly accelerated rectilinear motion, the acceleration of theparticle is constant. You may recognize these constant acceleration equationsfrom your physics courses.
0
0
0
constant
v t
v
dva dv a dt v v at
dt= = = =
0
210 0 0 0 2
0
x t
x
dxv at dx v at dt x x v t at
dt= = =
0 0
2 2
0 0constant 2
v x
v x
dvv a vdv a dx v v a x x
dx= = = =
These only apply to uniformly accelerated rectilinear motion…
Motion of Several Particles
Motion of several different particles, whose motion may beindependent or linked together.
Motion of Several Particles: Relative Motion
• For particles moving along the same line, time should be recorded from the same starting instant and displacements should be measured from the same origin in the same direction.
== ABAB xxx relative position of B with respect to A
ABAB xxx =
== ABAB vvv relative velocity of B with respect to A
ABAB vvv =
== ABAB aaa relative acceleration of Bwith respect to A
ABAB aaa =
Problem
Ball thrown vertically from 12 mlevel in elevator shaft with initialvelocity of 18 m/s. At same instant,open-platform elevator passes 5 mlevel moving upward at 2 m/s.
Determine (a) when and where ball hits elevator and (b) relative velocity of ball and elevator at contact.
SOLUTION:
• Substitute initial position and velocityand constant acceleration of ball intogeneral equations for uniformlyaccelerated rectilinear motion.
• Substitute initial position and constantvelocity of elevator into equation foruniform rectilinear motion.
• Write equation for relative position ofball with respect to elevator and solvefor zero relative position, i.e., impact.
• Substitute impact time into equationfor position of elevator and relativevelocity of ball with respect toelevator.
SOLUTION:
• Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion.
2
2
221
00
20
s
m905.4
s
m18m12
s
m81.9
s
m18
ttattvyy
tatvv
B
B
==
==
• Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion.
ttvyy
v
EE
E
==
=
s
m2m5
s
m2
0
• Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact.
025905.41812 2 == ttty EB
s65.3
smeaningles s39.0
=
=
t
t
• Substitute impact time into equations for position of elevator and relative velocity of ball with respect to elevator.
65.325=Ey
m3.12=Ey
65.381.916
281.918
=
= tv EB
s
m81.19=EBv
Motion of Several Particles: Dependent Motion
• Position of a particle may depend on position of one or more other particles.
• Position of block B depends on position of block A. Since rope is of constant length, it follows that sum of lengths of segments must be constant.
= BA xx 2 constant (one degree of freedom)
• Positions of three blocks are dependent.
= CBA xxx 22 constant (two degrees of freedom)
• For linearly related positions, similar relations hold between velocities and accelerations.
022or022
022or022
==
==
CBACBA
CBACBA
aaadt
dv
dt
dv
dt
dv
vvvdt
dx
dt
dx
dt
dx
Problem
Pulley D is attached to a collar which is pulled down at 75 mm/s. At t = 0, collar A starts moving down from K with constant acceleration and zero initial velocity. Knowing that velocity of collar A is 300 mm/s as it passes L, determine the change in elevation, velocity, and acceleration of block Bwhen block A is at L.
SOLUTION:
• Define origin at upper horizontal surfacewith positive displacement downward.
• Collar A has uniformly acceleratedrectilinear motion. Solve for accelerationand time t to reach L.
• Pulley D has uniform rectilinear motion.Calculate change of position at time t.
• Block B motion is dependent on motionsof collar A and pulley D. Write motionrelationship and solve for change of blockB position at time t.
• Differentiate motion relation twice todevelop equations for velocity andacceleration of block B.
SOLUTION:
• Define origin at upper horizontal surface with positive displacement downward.
• Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach L.
( )0
2
mm mm300 225 1.333 s
s s
= +
= =
A A Av v a t
t t
22
0 0
2
2
2
mm mm300 2 200mm 225
s s
A A A A A
A A
v v a x x
a a
=
= =
• Pulley D has uniform rectilinear motion. Calculate change of position at time t.
xD = xD( )0+ vDt
xD - xD( )0= 75
mm
s
æ
èç
ö
ø÷ 1.333s( ) =100 mm
• Block B motion is dependent on motions of collar A and pulley D. Write motion relationship and solve for change of block B position at time t.
Total length of cable remains constant,
0 0 0
0 0 0
0
2 2
2 0
200mm 2 100mm 0
A D B A D B
A A D D B B
B B
x x x x x x
x x x x x x
x x
=
=
=
( )0
400mmB Bx x- = -
• Differentiate motion relation twice to develop equations for velocity and acceleration of block B.
2 constant
2 0
mm mm300 2 75 0
s s
A D B
A D B
B
x x x
v v v
v
=
=
=
mm mm450 450
s sBv = - =
2
2 0
mm225 2(0) 0
s
A D B
B
a a a
a
=
=
2 2
mm mm225 225
s sBa = - =
CURVILINEAR MOTION
Curvilinear Motion: Position, Velocity & Acceleration
The softball and the car both undergo curvilinear motion.
A particle moving along a curve other than a straight line is in curvilinear motion.
• The position vector of a particle at time t is defined by a vector between origin O of a fixed reference frame and the position occupied by particle.
• Consider a particle which occupies position P defined by at time t and P’defined by at t + Dt,
r
r
Curvilinear Motion: Position, Velocity & Acceleration
Curvilinear Motion: Position, Velocity & Acceleration
0limt
s dsv
t dtD
D= =
D
Instantaneous velocity (vector)
Instantaneous speed (scalar)
0limt
r drv
t dtD
D= =
D
Curvilinear Motion: Position, Velocity & Acceleration
0limt
v dva
t dtD
D= = =
Dinstantaneous acceleration (vector)
• Consider velocity of a particle at time t and velocity at t + Dt,v
v
• In general, the acceleration vector is not tangent to the particle path and velocity vector.
Rectangular Components of Velocity & Acceleration
• When position vector of particle P is given by its rectangular components,
kzjyixr
=
• Velocity vector,
kvjviv
kzjyixkdt
dzj
dt
dyi
dt
dxv
zyx
=
==
• Acceleration vector,
kajaia
kzjyixkdt
zdj
dt
ydi
dt
xda
zyx
=
==2
2
2
2
2
2
Rectangular Components of Velocity & Acceleration
• Rectangular components particularly effective when component accelerations can be integrated independently, e.g., motion of a projectile,
00 ====== zagyaxa zyx
with initial conditions,
0,,0 000000 ==== zyx vvvzyx
Integrating twice yields
0
0
221
00
00
===
===
zgtyvytvx
vgtvvvv
yx
zyyxx
• Motion in horizontal direction is uniform.
• Motion in vertical direction is uniformly accelerated.
• Motion of projectile could be replaced by two independent rectilinear motions.
PROJECTILE MOTION
KINEMATIC EQUATIONS: HORIZONTAL MOTION
Since ax = 0, the velocity in the horizontal direction remains constant (vx = vox) and the position in the x direction can be determined by:
x = xo + (vox) t
Why is ax equal to zero (assuming movement through the air)?
KINEMATIC EQUATIONS: VERTICAL MOTION
Since the positive y-axis is directed upward, ay = – g. Application of the constant acceleration equations yields:
vy = voy – g t
y = yo + (voy) t – ½ g t2
vy2 = voy
2 – 2 g (y – yo)
For any given problem, only two of these three equations can be used. Why?
Problem
A projectile is fired from the edgeof a 150-m cliff with an initialvelocity of 180 m/s at an angle of30°with the horizontal. Neglectingair resistance, find (a) thehorizontal distance from the gun tothe point where the projectilestrikes the ground, (b) the greatestelevation above the groundreached by the projectile.
SOLUTION:
• Consider the vertical and horizontal motion separately (they are independent)
• Apply equations of motion in y-direction
• Apply equations of motion in x-direction
• Determine time t for projectile to hit the ground, use this to find the horizontal distance
• Maximum elevation occurs when vy=0
SOLUTION:
Given: (v)o =180 m/s (y)o =150 m
(a)y = - 9.81 m/s2 (a)x = 0 m/s2
Vertical motion – uniformly accelerated:
Horizontal motion – uniformly accelerated:
Choose positive x to the right as shown
SOLUTION:
Horizontal distance
Projectile strikes the ground at:
Solving for t, we take the positive root
Maximum elevation occurs when vy=0
Substitute into equation (1) above
Substitute t into equation (4)
Maximum elevation above the ground =
EXERCISE
Given: Projectile is fired with vA=150 m/s at point A.
Find: The horizontal distance it travels (R) and the time in the air.
Solving for tAB first, tAB = 19.89 s.
Then, R = 120 tAB = 120 (19.89) = 2387 m
Solution:
1) Place the coordinate system at point A. Then, write the equation for horizontal motion.
+ xB = xA + vAx tAB
where xB = R, xA = 0, vAx = 150 (4/5) m/s
Range, R will be R = 120 tAB
2) Now write a vertical motion equation. Use the distance equation.+ yB = yA + vAy tAB – 0.5 g tAB
2
where yB = – 150, yA = 0, and vAy = 150(3/5) m/s
We get the following equation: –150 = 90 tAB + 0.5 (– 9.81)tAB2
Motion Relative to a Frame in Translation
A soccer player must considerthe relative motion of the balland her teammates whenmaking a pass.
It is critical for a pilot toknow the relative motionof his aircraft with respectto the aircraft carrier tomake a safe landing.
Motion Relative to a Frame in Translation
• Designate one frame as the fixed frame of reference. All other frames not rigidly attached to the fixed reference frame are moving frames of reference.
• Position vectors for particles A and B with respect to the fixed frame of reference Oxyz are . and BA rr
• Vector joining A and B defines the position of Bwith respect to the moving frame Ax’y’z’ and
ABr
ABAB rrr
=
• Differentiating twice,
=ABv
velocity of B relative to A.ABAB vvv
=
=ABa
acceleration of B relative to A.
ABAB aaa
=
• Absolute motion of B can be obtained by combining motion of A with relative motion of B with respect to moving reference frame attached to A.
Problem
Automobile A is traveling east at theconstant speed of 36 km/h. As automobileA crosses the intersection shown,automobile B starts from rest 35 m northof the intersection and moves south with aconstant acceleration of 1.2 m/s2.Determine the position, velocity, andacceleration of B relative to A 5 s after Acrosses the intersection.
SOLUTION:
• Define inertial axes for the system
• Determine the position, speed, and acceleration of car A at t = 5 s
• Using vectors (Eqs 11.31, 11.33, and 11.34) or a graphical approach, determine the relative position, velocity, and acceleration
• Determine the position, speed, and acceleration of car B at t = 5 s
Problem
SOLUTION: • Define axes along the road
Given: vA=36 km/h, aA= 0, (xA)0 = 0
(vB)0= 0, aB= - 1.2 m/s2, (yA)0 = 35 m
Determine motion of Automobile A:
We have uniform motion for A so:
At t = 5 s
11 - 48
SOLUTION:
Determine motion of Automobile B:
We have uniform acceleration for B so:
At t = 5 s
SOLUTION:
We can solve the problems geometrically, and apply the arctangent relationship:
Or we can solve the problems using vectors to obtain equivalent results:
= B A B/ Ar r r = B A B/Av v v = B A B/Aa a a
20 50
20 50 (m)
=
=
B/A
B/A
j i r
r j i
6 10
6 10 (m/s)
=
=
B/A
B/A
j i v
v j i 2
1.2 0
1.2 (m/s )
=
=
B/A
B/A
j i a
a j
Physically, a rider in car A would “see” car B travelling south and west.
TANGENT AND NORMAL COMPONENTS
Tangential and Normal Components
If we have an idea of the path of a vehicle, it is often convenient to analyze the motion using tangential and
normal components (sometimes called path coordinates).
Normal and Tangential Components
When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known, normal (n) and tangential (t) coordinates are often used.
In the n-t coordinate system, the origin is located on the particle (the origin moves with the particle).
The t-axis is tangent to the path (curve) at the instant considered, positive in the direction of the particle’s motion.The n-axis is perpendicular to the t-axis with the positive direction toward the center of curvature of the curve.
The position of the particle at any instant isdefined by the distance, s, along the curve from a fixed reference point.
The positive n and t directions are defined by theunit vectors un and ut, respectively.
The center of curvature, O’, always lies on theconcave side of the curve.The radius of curvature, r, is defined as theperpendicular distance from the curve to thecenter of curvature at that point.
VELOCITY IN THE n-t COORDINATE SYSTEM
The velocity vector is always tangentto the path of motion (t-direction).
The magnitude is determined by taking the time derivative of the path function, s(t).v = v ut where v = s = ds/dt
Here v defines the magnitude of the velocity (speed) andut defines the direction of the velocity vector.
ACCELERATION IN THE n-t COORDINATE SYSTEM
Acceleration is the time rate of change of velocity:a = dv/dt = d(vut)/dt = vut + vut
. .
Here v represents the change in the magnitude of velocity and ut represents the rate of change in the direction of ut.
.
.
.a = v ut + (v2/r) un = at ut + an un.
After mathematical manipulation, the acceleration vector can be expressed as:
ACCELERATION IN THE n-t COORDINATE SYSTEM (continued)
So, there are two components to the accelerationvector:
a = at ut + an un
• The normal or centripetal component is always directed toward the center ofcurvature of the curve. an = v2/r
• The tangential component is tangent to the curve and in the direction of increasingor decreasing velocity.
at = v or at ds = v dv.
• The magnitude of the acceleration vector isa = [(at)
2 + (an)2]0.5
SPECIAL CASES OF MOTIONThere are some special cases of motion to consider.
1) The particle moves along a straight line.r => an = v2/r = 0 => a = at = v
The tangential component represents the time rate of changein the magnitude of the velocity.
2) The particle moves along a curve at constant speed.at = v = 0 => a = an = v2/r
.
The normal component represents the time rate of change inthe direction of the velocity.
3) The tangential component of acceleration is constant, at = (at)c.
In this case,
s = so + vo t + (1/2) (at)c t2
v = vo + (at)c t
v2 = (vo)2 + 2 (at)c (s – so)
As before, so and vo are the initial position and velocity of the particle at t = 0. Howare these equations related to projectile motion equations? Why?
4) The particle moves along a path expressed as y = f(x).The radius of curvature, r, at any point on the path can be calculated from
r = ________________]3/2(dy/dx)21[
2d2y/dx
Problem
A motorist is traveling on a curvedsection of highway of radius 750 m atthe speed of 90 km/h. The motoristsuddenly applies the brakes, causing theautomobile to slow down at a constantrate. Knowing that after 8 s the speedhas been reduced to 72 km/h,determine the acceleration of theautomobile immediately after thebrakes have been applied.
SOLUTION:
• Define your coordinate system
• Calculate the tangential velocity and tangentialacceleration
• Determine overall acceleration magnitudeafter the brakes have been applied
• Calculate the normal acceleration
21.041m/s=a 53.1a =
2
2
0.833m/stan
0.625m/s
n
t
a
aa = =
2 22(25m/s)
0.833m/s750 m
n
va
r= = =
220 m/s 25m/saverage 0.625m/s
8 st t
va a
t
D -= = = = -
D
km 1000 m 1 h90 km/h = 90 25 m/s
h 1 km 3600 s
72 km/h = 20 m/s
=
SOLUTION: • Define your coordinate system
eten
• Determine velocity and acceleration in thetangential direction
• The deceleration constant, therefore
• Immediately after the brakes are applied,the speed is still 25 m/s
2 2 2 2( 0.625) (0.833)n ta a a= =
Reference
Vector Mechanics for Engineers Statics and Dynamics: Ferdinand P. Beer; Elwood Russell Johnston ninth edition
Collision of elastic bodiesWe will also see that if the balls of different materials are allowed to fall on a marble floor, they will rebound to different Heights. This property of bodies, by virtue of which, they rebound, after impact, is called “elasticity”. It may be noted that a body, which rebounds to a greater height is said to be more elastic, than that which rebounds to a lesser height. But, if a body does not rebound at all, after its impact, it is called an “inelastic body”.
PHENOMENON OF COLLISION
Whenever two elastic bodies collide with each other, the phenomenon of collision takes place as given below :
1. The bodies, immediately after collision, come momentarily to rest.2. The two bodies tend to compress each other, so long as they are compressed to the maximum value.3. The two bodies attempt to regain its original shape due to their elasticity. This process ofregaining the original shape is called restitution.
The time taken by the two bodies in compression, after the instant of collision, is called the“Time of compression” and time for which restitution takes place is called the “Time of restitution”. The sum of the two times of collision and restitution is called time of collision, period of collision, or period of impact
LAW OF CONSERVATION OF MOMENTUM
It states, “The total momentum of two bodies remains constant after their collision or any other mutual action.”
Mathematically
where m1 = Mass of the first body,
u1 = Initial velocity of the first body,
v1 = Final velocity of the first body, and
m2, u2, v2 = Corresponding values for the second body
NEWTON’S LAW OF COLLISION OF ELASTIC BODIES
It states, “ When two moving bodies collide with each other, their velocity of separation bears
a constant ratio to their velocity of approach.”
Mathematically,
where e is a constant of proportionality, and is called the “coefficient of restitution”. Its value lies between 0 and 1.
It may be noted that if e = 0, the two bodies are inelastic. But if e = 1, the two bodies are perfectly elastic.
TYPES OF COLLISIONS
When two bodies collide with one another, they are said to have an impact. Following are the
two types of impacts.
1. Direct impact, and 2. Indirect (or oblique) impact.
The line of impact, of the two colliding bodies, is the
line joining the centre of these bodies and passes through the
point of contact or point of collision.
If the two bodies, before impact, are moving along the
line of impact, the collision is called as Direct impact.
LOSS OF KINETIC ENERGY DURING COLLISION
The loss of kinetic energy, during impact, may be obtained by finding out the kinetic energy of the two bodies before and after the impact. The difference between the kinetic energies of the system, gives the required loss of kinetic energy during impact. Consider two bodies A and B having a direct impact.
INDIRECT IMPACT OF TWO BODIES
If the two bodies, before impact, are not moving along the line of
impact, the collision is called an Indirect (or oblique) impact.
A little consideration will show, that the components of u1, u2, v1 and v2 along the line of impact, will cause the
direct impact and all the relations for direct impact, will hold good for these components. The other components of
u1, u2, v1 and v2, ( i. e., along the perpendicular to the line of impact) will not effect the phenomenon of impact in
any way, and the component of initial velocity will be equal to the component of final velocity of body.
INDIRECT IMPACT OF A BODY WITH A FIXED PLANE
NUMERICAL PROBLEM
Q.1 A ball of mass 1 kg moving with a velocity of 2 m/s impinges directly on a ball of mass 2 kg at rest. The first ball , after impinging , comes to rest. Find the velocity of the second ball after the impact and the coefficient of restitution.
Q.2 Three perfectly elastic balls A, B and C of masses 2 kg, 4 kg and 8 kg move in the same direction with velocities of 4 m/s, 1m/s and 0.75 m/s respectively. If the ball A impinges with the ball B, which in turn, impinges with the ball C, prove that the balls A and B will be brought to rest by the impacts.
Q.3 A ball impinges directly on a Similar ball at rest. The first ball is reduced to rest by the impact. Find the coefficient of restitution , if half of the initial kinetic energy is lost by impact.
A ball of mass 2 kg, moving with a velocity of 3 m/sec, impinges on a ball of mass 4 kg moving with a velocity of 1 m/sec. The velocities of two balls are parallel and inclined at 30° to the line joining their centres at the instant of impact. If the coefficient of restitution be 0.5, find( a ) direction , in which the 4 kg ball will move after impact ;( b ) velocity of the 4 kg ball after impact ;( c) direction , in which the 2 kg ball will move after impact ; and( d ) velocity of the 2 kg ball after impact.
Reference:- (1) Mechanics oif solid by S.S Bhavikatti. (2) Engineering Mechanics by R.S KhurmiReference:- (1) Mechanics oif solid by S.S Bhavikatti. (2) Engineering Mechanics by R.S Khurmi
