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Impulse Momentum
It is clear from the discussion of the previous chapters that for solving kinetic problems, involving force and acceleration, D’Alembert’s principle is useful and that for the problems involving force, velocity and displacement, the work-energy method is useful. In this chapter, the Impulse Momentum method is dealt which is useful for solving the problems involving force, time and velocity.
17.1. LINEAR IMPULSE AND MOMENTUM
If R is the resultant force acting on a body of mass m , then from Newton’s second law, R = ma But acceleration a = dv / dt
∴ R = m dv dt i.e., Rdt = mdv
∴ z Rdt = z mdv
If initial velocity is u and after time interval t it becomes v , then
0
t
z Rdt^ =^ m^ v^ u
v = mv – mu ...(17.1)
The term (^0)
t
z R^ ·^ dt^ is called^ impulse. If the resultant force is in newton and time is in
second, the unit of impulse will be N-sec. If R is constant during time interval t , then impulse is equal to R × t. The term mass × velocity is called momentum.
Now, mv =
W
g v
Substituting dimensional equivalence, we get,
=
N
m/sec 2
m/sec = N-sec
Thus the momentum has also unit N-sec. The equation (17.1) satisfies the requirement of dimensional homogenity. Equation (17.1) can now be expressed as
396 FUNDAMENTALS OF ENGINEERING MECHANICS
Impulse = Final momentum – Initial momentum ...(17.2) Since the velocity is a vector, impulse is also a vector. The impulse momentum equation (eqn. 17.1 or 17.2) holds good when the directions of R , u and v are the same. Impulse momen- tum equation can be stated as follows: The component of the resultant linear impulse along any direction is equal to change in the component of momentum in that direction. The impulse momentum equation can be applied in any convenient direction and the kinetic problems involving force, velocity and time can be solved easily. Example 17.1. A glass marble, whose weight is 0.2 N, falls from a height of 10 m and rebounds to a height of 8 metres. Find the impulse and the average force between the marble and the floor, if the time during which they are in contact is 1/10 of a second. Solution. Applying kinematic equations, for the freely falling body (Ref. Fig. 17.1), the velocity with which marble strikes the floor
= (^2) gh = (^2) × 9 81. × 10 = 14.007 m/sec (downward) ...(1) Similarly applying kinematic equations for the marble mov- ing up, we get the velocity of rebound
= 2 g × 8 = 2 × 9 81. × 8 = 12.528 m/sec (upward) ...(2) Taking upward direction as positive and applying impulse momentum equation, we get
Impulse =
W
g ( v – u )
= 0 2 9 81
[12.52 – (– 14.007)]
= 0.541 N-sec Ans. If F is the average force, then Ft = 0. F × 1/10 = 0.541 N F = 5.41 N Ans. Example 17.2. A 1 N ball is bowled to a batsman. The velocity of ball was 20 m/sec horizontally just before batsman hit it. After hitting it went away with a velocity of 48 m/sec at an inclination of 30° to horizontal as shown in Fig. 17.2(a). Find the average force exerted on the ball by the bat if the impact lasts for 0.02 sec.
48 m/sec
30° q
u = 20 m/sec F (^) y
F (^) x
P
F (^) y
F (^) x
(a) (b) Fig. 17.2( a , b )
Fig. 17.
A
B
10 m10 m 8 m8 m
398 FUNDAMENTALS OF ENGINEERING MECHANICS
Example 17.4. A 20 kN-automobile is moving at a speed of 70 kmph when the brakes are fully applied causing all four wheels to skid. Determine the time required to stop the automo- bile (a) on concrete road for which μ = 0.75, (b) on ice for which μ = 0.08. Solution. Initial velocity of the vehicle
u = 70 kmph =
×
×
= 19.44 m/sec Final velocity v = 0 Free body diagram is shown in Fig. 17. F = μ N = μ W = 20 μ Applying impulse momentum equation, we have
W
g ( v – u )
–20 μ t =
L
NM^
O
QP^
∴ t =
μ ( i ) On concrete road μ = 0.
∴ t =
= 2.64 sec Ans. ( ii ) On ice, μ = 0.
∴ t =
= 24.78 sec Ans. Example 17.5. A block weighing 130 N is on an incline, whose slope is 5 vertical to 12 horizontal. Its initial velocity down the incline is 2.4 m/sec. What will be its velocity 5 sec. later? Take coefficient of friction at contact surface = 0.3.
Solution. tan θ =
⇒ θ = 22.62° N = W cos θ = 130 cos 22.62° = 120 newton F = μ N = 0.3 × 120 = 36 newton ∑ Forces down the plane = R = W sin θ – F = 130 sin 22.26° – 36 = 14.0 N Initial velocity u = 2.4 m/sec. Let final velocity be v m/sec. Time interval t = 5 sec. Applying impulse momentum equation,
Rt =
W
g ( v – u ), we get
14 × 5 =
L
NM^
O
QP^
( v – 2.4) ∴ v = 7.68 m/sec Ans.
Fig. 17.
Fig. 17.
W
F N
q
130N
N
F
12
5
IMPULSE MOMENTUM 399
17.2. CONNECTED BODIES
The problems involving connected bodies may be solved by any one of the following two methods: First Method: Free body diagrams of each body is drawn separately. Impulse momen- tum equation for each body in the direction of its motion is written and then the equations are solved to get the required values. Second Method: If the connected bodies have same displacement in the same time, the impulse of internal tension in connecting chords will get cancelled. Hence free body diagram of combined bodies may be considered and impulse moment equation applied in the direction of motion of combined bodies. This method is applicable only if displacement of each body is the same in given time. Example 17.6. Determine the time required for the weights shown in Fig. 17.6(a) to attain a velocity of 9.81 m/sec. What is the tension in the chord? Take μ = 0.2 for both planes. Assume the pulleys as frictionless. Solution. First Method: Free body diagram of the two blocks are as shown in Fig. 17.6( b ). For 2000 N block: N 1 = W 1 cos 30° = 2000 cos 30° = 1732.05 N F 1 = μ N 1 = 0.2 × 1732. = 346.41 N For 1800 N block: N 2 = W 2 cos 60° = 1800 cos 60° = 900 N F 2 = μ N 2 = 0.2 × 900 = 180 N.
2000N
30° 60°
1800N
W 1 = 2000N T
F (^1) N 1 N 2
F (^2)
T
W 2 = 1800N
(a) (b)
W 1 = 2000N W^2 = 1800N
F (^2)
F (^1) N 1
N 2
(c) Fig. 17.
IMPULSE MOMENTUM 401
Example 17.7. Determine the tension in the strings and the velocity of 1500 N block shown in Fig. 17.7(a) 5 seconds after starting from ( a ) rest ( b ) starting with a downward velocity of 3 m/sec. Assume pulleys as weightless and frictionless.
1500N
500N 1500N
2T 500N
T T T
Fig. 17. Solution. When 1500 N block moves a distance s , in the same time 500 N block moves a distance 2 s. Hence if velocity of 1500 N block is v m/sec that of 500 N block will be 2 v m/sec. Let T be the tension in the chord connecting 500 N block. Hence tension in the wire connecting 1500 N block will be 2 T (see Fig. 17.7( b )). Since the velocities of two blocks are different only first method is to be used. Case ( a ) : Initial velocity u = 0, t = 5 sec. Writing impulse momentum equation for 500 N block, we have
( T – 500) t =
(2 v – u )
( T – 500)5 =
× (2 v – 0)
T – 500 =
v .
Applying impulse momentum equation to 1500 N block,
(1500 – 2 T ) t =
( v – 0)
(1500 – 2 T ) 5 =
v
1500 – 2 T =
v .
402 FUNDAMENTALS OF ENGINEERING MECHANICS
Adding Eqn. (2) in 2 times Eqn. (1), we get
1500 – 2 × 500 =
v v v
...
∴ v = 7.007 m/sec Ans. Substituting it is Eqn. (1), we get
T – 500 =
× 7.
∴ T = 642.86 N Ans. Case ( b ) : Initial velocity u = 3 m/sec. Impulse momentum equation for 500 N body will be
( T – 500) 5 = 500 9 81.
(2 v – 3)
i.e., T – 500 =
v − ...(3)
Impulse momentum equation for 1500 N body will be
(1500 – 2 T ) 5 =
( v – 3)
i.e., 1500 – 2 T = 300 9 81.
( v – 3) ...(4)
Adding Eqn. (4) and 2 times Eqn. (1), we get
1500 – 1000 =
(7 v – 15)
v = 9.15 m/sec Ans. Substituting it in Eqn. (3), we get
T – 500 = 100 9 81.
(2 × 9.15 – 3)
∴ T = 655.96 N Ans. Example 17.8. The system shown in Fig. 17.8(a) has a rightward velocity of 3 m/sec. Determine its velocity after 5 seconds. Take μ = 0.2 for the surfaces in contact. Assume pulleys to be frictionless.
Solution. Since all bodies have same displacement in given time, consider the combined FBD of the system.
N 1 = 500 N F 1 = 0.2 × 500 = 100 N N 2 = 1000 cos 30° = 866. F 2 = 0.2 × N 2 = 173.2 N
404 FUNDAMENTALS OF ENGINEERING MECHANICS
Applying impulse momentum equation for the motion from stationary position to leftward motion, after total time of 20 sec, we have
( P – F – 400) (20 – t 1 ) = 1000 400 9 81
( v – 0)
( P – 600) (20 – t 1 ) =
× 6 ...(2)
Simultaneous equations (1) and (2) may be solved to get t 1 and P. Trial and error method may be advantageously used here. Looking at Eqn. (2), the value of P should be more than 600. Let us take a trial value of P as 700. From Eqn. (1) t 1 = 1.142 sec Substituting it in Eqn. (2), we get P = 645.41 N Substituting this value of P in Eqn. (1), we get t 1 = 1.282 sec Substituting it in Eqn. (2), we get P = 645.74 N This value is almost same as trial value 646.41 N Hence, P = 645.74 N Ans.
17.3. FORCE OF JET ON A VANE
In hydroelectric generating stations, a jet of water is made to impinge on the vanes of turbines and get deflected by a certain angle. During this process a force is exerted by the jet on the vane and that causes rotation of turbine. This machanical energy is further converted into electric energy. The force exerted by the jet on the vane, moving or stationary, can be determined by applying impulse momentum equations. This is illustrated with examples 17. and 17.11. Example 17.10. A nozzle issues a jet of water 50 mm in diameter, with a velocity of 30 m/sec which impinges tangentially upon a perfectly smooth and stationary vane, and deflects it through an angle of 30° without any loss of velocity (see Fig. 17.10). What is the total force exerted by the jet upon the vane? Solution. Weight of water whose momentum is changed in t second = (π/4) (0.05) 2 × 30 × 9810 × t = 577.86 t newtons. ( Note: 1 cubic metre of water weighs 9810 newton) Let P (^) x and Py be the components of reactive force of vane.
Nozzle JetJet 30 m/sec Vane P^ y
30°Px
30 m/sec
Fig. 17.
IMPULSE MOMENTUM 405
Applying impulse momentum equation in x -direction, we get
t (^) (30 cos 30° – 30)
P (^) x = 236.75 N Applying the impulse momentum equation in y -direction
P (^) y t = 577 86 9 81
t (^) (30 sin 30° – 0)
P (^) y = 883.58 N P = (^) Px^2^ + Py^2 = 236 75. 2 +883 58.^2 = 914.75 N Inclination with horizontal,
θ = tan –^
P
P
y x
F
HG^
I
KJ^
= tan–^
θ = 75.0° The force P shown in Figure 17.11( a ) is the reactive force of the vane. The force of jet is equal and opposite to this force as shown in Fig. 17.11( b ).
q
P
75°
P (a) (b) Fig. 17. Example 17.11. In the previous example if the vane is moving with a velocity of 10 m/sec towards right, what will be the pressure exerted by the jet. Solution. Velocity of approach = 30 – 10 = 20 m/sec Weight of water impinging in t second = (π/4) × (0.05)^2 × 20 × t × 9810 = 385.24 t Velocity of departure = velocity of approach = 30 – 10 = 20 m/sec Writing impulse moment equation in x direction
t (^) (20 cos 30° – 20)
P (^) x = 105.22 N
IMPULSE MOMENTUM 407
Example 17.12. A 800 N man, moving horizontally with a velocity of 3 m/sec, jumps off the end of a pier into a 3200 N boat. Determine the horizontal velocity of the boat (a) if it had no initial velocity and (b) if it was approaching the pier with an initial velocity of 0.9 m/sec. Solution. Weight of man W 1 = 800 N Velocity with which man is running v = 3 m/sec. Weight of the system after man jumps into boat = 800 + 3200 = 4000 N. ( a ) Initial velocity of boat = 0 Since the action of the man is equal to the reaction of the boat, the principle of conservation of momentum can be applied to the system consisting of the man and the boat. Initial momentum = Final momentum 800 9 81
× + × = v
v = 0.6 m/sec Ans. ( b ) Initial velocity of boat = 0.9 m/sec towards the pier = –0.9 m/sec Applying principle of conservation of momentum, we get 800 9 81
× + × − = v
∴ v = –0.12 m/sec i.e., velocity of boat and man will be 0.12 m/sec towards the pier. Ans. Example 17.13. A car weighing 11,000 N and running at 10 m/sec holds three men each weighing 700 N. The men jump off from the back end gaining a relative velocity of 5 m/sec with the car. Find the speed of the car if the three men jump off ( i ) in succession, ( ii ) all together. Solution. ( i ) When three men jump off in succession initial velocity u = 10 m/sec. Let the velocity when ( a ) first man jumps be v 1 m/sec ( b ) second man jumps be v 2 m/sec ( c ) third man jumps be v 3 m/sec. Velocity of the first man w.r.t. fixed point when he jumps = v 1 – 5. Applying principle of conservation of momentum when the first man jumps, we get (11,000 + 3 × 700) 10 = (11,000 + 2 × 700) v 1 + 700 ( v 1 – 5) = (11,000 + 3 × 700) v 1 – 700 × 5
v 1 = 10 + 700 5 11 000 3 700
×
, + ×
When the second man jumps (11,000 + 2 × 700) v 1 = (11,000 + 700) v 2 + 700( v 2 – 5)
v 2 = v 1 +
×
, + ×
408 FUNDAMENTALS OF ENGINEERING MECHANICS
When the third man jumps: (11,000 + 700) v 2 = 11,000 v 3 + 700 ( v 3 – 5)
v 3 = v 2 +
×
From (1), (2) and (3), we get
v 3 = 10 + 700 × 5
, + × , 11 000, 700
+ ×
F
HG^
I
KJ
= 10.849 m/sec Ans. ( ii ) When three men jump together Let the velocity of the car be v when three men jump together. Applying principle of conservation of momentum, we get (11,000 + 3 × 700) 10 = 11,000 v + 3 × 700( v – 5) where, ( v – 5) is the relative velocity of the men when they jump,
v = 10 +
× ×
, + ×
i.e. , v = 10.802 m/sec Ans. Example 17.14. A car weighing 50 kN and moving at 54 kmph along the main road collides with a lorry of weight 100 kN which emerges at 18 kmph from a cross road at right angles to main road. If the two vehicles lock after collision, what will be the magnitude and direction of the resulting velocity?
18 kmph
50 kN 100 kN
54 kmph
v (^) y
vx^12
v
18
q
(a) (b) Fig. 17. Solution. Let the velocity of the vehicles after collision be v (^) x in x direction (along main road) and v (^) y in y direction (along cross road) as shown in Fig. 17.13( a ). Applying impulse moment equation along x direction, we get
50 54 9 81
×
v (^) x ∴ v (^) x = 18 kmph Applying impulse momentum equation in y direction, we get
0 +
×
v y v (^) y = 12 kmph
410 FUNDAMENTALS OF ENGINEERING MECHANICS
( Note: Work done by the weight is negative since weight is a force acting downwards force whereas body has moved upwards).
u = 1.025 m/sec Let v be the velocity of bullet before striking the block. Applying principle of conserva- tion of momentum to the bullet and block system, we get
0 3 9 81
v + = +^ u
0.3 v = 100.3 × 1. v = 342.69 m/sec Ans.
Initial energy of bullet = 0 3 2 9 81
×.
(342.69) 2 = 1795.68 J
Energy of the block and bullet system
= 1 2
× +^.
1.025 2 = 5.37 J
Loss of energy = 1795.68 – 5. = 1790.31 J Ans. Example 17.17. A bullet weighs 0.5 N and moving with a velocity of 400 m/sec hits centrally a 30 N block of wood moving away at 15 m / sec and gets embedded in it. Find the velocity of the bullet after the impact and the amount of kinetic energy lost.
400 m/sec 0.5 N
15 m/sec
30 N Fig. 17. Solution. Initial momentum of the system = Final momentum 0 5 9 81
× 400 + 30
× = +^ v
v = 21.31 m/sec Kinetic Energy lost = Initial K.E. – Final K.E.
=
F × × 2 + × 2
HG^
I
KJ^
. − ×
× 21.31 2
= 3715.47 J Ans.
17.5. PILE AND PILE HAMMER
If the safe bearing capacity of the soil is too less, a set of reinforced concrete or steel poles are driven in the soil. Such poles are known as piles. Over the group of piles concrete cap is cast and on it the structure is built.
IMPULSE MOMENTUM 411
The piles are driven by pile hammer. It consists of a movable weight called the hammer (see Fig. 17.16). The hammer is raised to a convenient height h and freely dropped. It is guided to fall over the pile. After the hammer strikes the pile the hammer and the pile move downward together. The kinetic energy of the pile and the hammer is utilised in doing the work against resistance of the ground and pile gets driven by a distance s. By repeated hammering the pile is driven to required depth. If the distance moved per blow is known, earth resistance can be calculated. A general equation is derived in Ex. 17.18, and then two specific examples are solved. Example 17.18. A pile of weight W is driven vertically through a distance s when a hammer of weight w is dropped from a height h. Calculate the average resistance of the ground, the loss of kinetic energy during the impact and the time during which the pile is in motion. Solution. Initial velocity of hammer = 0 Distance moved before striking pile = h m Gravitational acceleration = g m/sec^2 ∴ Velocity of hammer while striking pile v is given by v^2 – u^2 = 2 gh v^2 – 0 = 2 gh v = 2 gh …(1) After striking, the hammer and the pile move together. Hence, the momentum is con- served. Applying the principle of conservation of momentum
wv g
w W g
V
where, V is the velocity of hammer and the pile immediately after the strike.
V = w w + W v …(2)
With this initial velocity the pile and hammer start moving downwards and they stop moving after a distance s. During this process, work is done against the resistance R of the earth. Applying work energy equation, we have
( w + W ) s – R × s =
w W g
(0 – V^2 )
Rs = ( w + W ) s + w W g
V^2 …(3)
From (2), V = w w + W v
∴ Rs = ( w + W ) s + w W g
w w W
2 ( ) 2 v
2
Fig. 17.
W h
W
SS
IMPULSE MOMENTUM 413
Applying the principle of conservation of momentum of pile and hammer, we get velocity V of the pile and hammer immediately after the impact.
20 9 81
× = +^ V
V = 20
× 3 836. = 2.557 m/sec
Applying work energy equation to the motion of the hammer and pile, resistance R of the ground can be obtained.
(20 + 10 – R ) s = 20 10 2
g
(0 – V^2 )
(30 – R ) 0.1 = 30
2 × 9 81.
R = 130 kN Ans. Example 17.20. A pile hammer, weighing 15 kN drops from a height of 600 mm on a pile of 7.5 kN. How deep does a single blow of hammer drive the pile if the resistance of the ground to pile is 140 kN?
Assume that ground resistance is constant. Solution. h = 600 mm = 0.6 m Velocity of hammer at the time of strike v = 2 gh = 2 × 9 81. ×0 6. = 3.431 m/sec Let V be the velocity of pile and hammer immediately after impact. Applying principle of conservation of momentum to the system of pile and pile hammer, we get
15 9 81.
× 3.431 + 0 = 15 7 5
V
V =
×.
= 2.287 m/sec
Now applying work energy equation to the system, we get
(15 + 7.5 – R ) s =
×
(0 – V^2 )
(15 + 7.5 – 140) s =
×
s = 22 5 2 9 81
×.
(+ 2.287^2 )
= 0.051 m
s = 51 mm Ans. Example 17.21. A hammer weighing 5 N is used to drive a nail of weight 0.2 N with a velocity of 5 m/sec. horizontally into a fixed wooden block. If the nail penetrates by 20 mm per blow, calculate the resistance of the block, which may be assumed uniform.
414 FUNDAMENTALS OF ENGINEERING MECHANICS
Solution. Applying impulse momentum equation to the system of hammer and nail, we get
5 9 81_._
× 5 + 0 = 5 0 2
V
V = 4.808 m/sec Applying work energy equation to the system shown in Fig. 17.17, we get
×.
R × 0.02 =
×.
× 4.808 2
R = 306.3 N Ans. Note: Weights of hammer and nail do not do any work since there is no displacement in the directions of these forces.
Important Definitions
1. The product of mass and velocity is called momentum. 2. If R is the resultant force acting on the body for an interval zero to t, then the term
0
t
z R dt is called^ impulse.
3. Impulse momentum equation can be stated as ‘the component of resultant linear impulse along any direction is equal to change in the component of momentum in that direction’.
Important Equations
0
t
z R dt =^ mv^ –^ mu.
2. If momentum is conserved in a system, W W g
1 +^2 v = W g
(^1) v 1 +^
W
g
(^2) v
PROBLEMS FOR EXERCISE
17.1 A cricket ball weighing one newton approaches a batsman with a velocity of 18 m/sec in the direction shown in Fig. 17.18. After hit by the bat at B , it moves out with a velocity of 40 m/sec at 45° to horizontal. If the bat and ball were in contact for 0.02 sec, determine the impulsive force exerted by the bat. [ Ans. 230.37 N]
Fig. 17.
V
5N
0.2N
R
