In this video concept of law of conservation of momentum, impact, types of impact with numerical is explained

1. Kinetics of Particles
(Impulse and Momentum)
Presented By,
Miss. Shinde Bharti M. (Assistant Professor)
Department of Civil Engineering
Sanjivani College of Engineering, Kopargaon.
Email- shindebharticivil@sanjivani.org.in
1

2. Contents-
1. Impulsive Force and Impulse
2. Momentum
3. Impulse Momentum Equation
4. Law of Conservation of Momentum
5. Impact
• Coefficient of Restitution
• Types of Impact
2

3. 3
Impulsive Force-
It is defined as the large force exerted in a short interval of time during
collision or explosion.
When two bodies collides, for a short interval of time, at a point of contact
action and reaction will exerts an impulsive forces.
Ex. Batsman hits the ball in cricket/baseball, a force exerted by a hammer on a
nail etc.
Impulse of Force (I)
It is the product of impulsive force and a time interval.
I = F x t
Where, F- is the impulsive force
t- is the time interval

4. 4 Momentum (M)-
It is the measure of a motion of a moving body expressed as,
M = mass x velocity
M = mv

5. 5
Impulse Momentum Equation-
It is also called as a principle of impulse and momentum.
According to Newton’s 2nd law,
2 2
1 1
2
1
2
1
t v
t v
t
2 1
t
t
t
F ma
dv
And,a acceleration
dt
dv
F m
dt
Fdt mdv
Fdt mdv
Fdt mv mv
Fdt impulseof Force
Impulse=finalmomentum-Initialmomentum

 
 
 
 
  





 



6. 6
Law of Conservation of Momentum-
From Principle of Impulse and Momentum,
I= Change in momentum
I= m*v
When the forces (action and reaction at the point of contact )exerted during
impact within a short interval of time, are same in magnitude but opposite in
direction, the sum of impulse is zero during the impact. (I=0)
Therefore, Change in momentum=0
Initial Momentum = Final momentum
m1V1+ m2V2 = m1V’1+ m2V’2
Where, m1 & m2 – mass of the objects 1 and 2 respectively
V1 & V2 -Velocities of the objects 1 and 2 respectively just before the impact
V’1 & V’2 – Velocities of the objects 1 and 2 respectively just after the impact

7. 7
Impact-
 Coefficient of Restitution (e)-
V1>V2 V’1<V’2
V1-V2= is the Velocity of Approach V’2-V’1= is the Velocity of Separation
Therefore, coefficient of restitution is the ratio of velocity of separation to the
velocity of approach.
e =
velocity of separation
velocity of approach
=
V′2−V′1
V1−V2
1 2
V1
V2
1 2
V’1
V’2

8. 8
Types of Bodies/Impact-
Based on the value of coefficient of restitution (e) the impact/ body can
classified into three categories as,
1. Perfectly Elastic Impact/Body- e = 1
2. Semi-elastic Impact/Body - 0 < e <1
3. Perfectly Plastic Impact/body - e = 0
e =
V′2−V′1
V1−V2
= 0
V′2−V′1=0 V′2=V′1=V′
From law of conservation of momentum,
m1V1+ m2V2 = m1V’1+ m2V’2
m1V1+ m2V2 = (m1+ m2)V’
V’ =
m1V1+ m2V2
(m1+ m2)

9. 9
Types of Impact-
1. Direct Central Impact-
If the velocities of the object 1 and 2 before the impact and after the
impact are collinear with the line of impact then it is called as Direct Central
Impact.
In case of direct central impact the equation of law of conservation of
momentum and coefficient of restitution are valid.
V2
V’1
V’2
Line of Impact

10. 10 Coefficient of Restitution-
In case of, direct central impact the formula for coefficient of restitution
can be re-write as,
where, h = rebound Height
H = Height of drop
h
e
H


11. 11
2. Indirect/Oblique Impact-
If the velocities of the object 1 and 2 before the impact and after the
impact are not collinear with the line of impact then it is called as
Indirect/Oblique Impact.
V2
V’1
V’2
Line of Impact

12. 12
Examples-
1. A 35 x 106 Kg ocean liner has an initial velocity of 4 km/h. Neglecting the
frictional resistance of the water, determine the time required to bring the liner
to rest by using a single tugboat which exerts a constant force of 150 kN.
Ans:
m = 35 x 106 Kg, F =150 x 103 N
v = 4 km/hr =1.1111 m/s
Using principle of impulse and momentum,
1
6 3
mv Ft 0
(35x10 x1.11) (150x10 )t 0
t 259.26s
 
  
 

13. 13
2. A sailboat weighing 4.5 KNwith its occupants is running down wind at 3.6 m/s
when its spinnaker is raised to increase its speed. Determine the net force
provided by the spinnaker over the 10-s interval that it takes for the boat to
reach a speed of 5.4 m/s.
Ans:
m = 4.5x 103 N , t= 10s
v1 = 3.6 m/s, v2 =5.4 m/s
Using principle of impulse and momentum,
1 2
3 3
mv Ft mv
(4.5x10 x3.6) (Fx10) (4.5x10 x5.4)
F 810N
 
  
 

14. 14
3. At an intersection car B was traveling south and car A was traveling 30° north of
east when they slammed into each other. Upon investigation it was found that
after the crash the two cars got stuck and skidded off at an angle of 10° north of
east. Each driver claimed that he was going at the speed limit of 50 km/h and
that he tried to slow down but couldn’t avoid the crash because the other driver
was going a lot faster. Knowing that the masses of cars A and B were 1500 kg
and 1200 kg, respectively, determine (a) which car was going faster, (b) the
speed of the faster of the two cars if the slower car was traveling at the speed
limit.

15. 15
Ans:
In this case, the total momentum of the cars is conserved.,
 
 
A A A B
A A B B A B
A B
B
A
mv,x : m v cos30 m m vcos10......................(1)
mv,y : m v sin30 m v m m vsin10..........(2)
solving eq(1) into eq(2),
v 2.30v
which shows that car A is moving faster than car B
v 50km / h
v 115km / h
 
  

 
 



16. 16
4. A 20-Mg railroad car moving at a speed of 0.5 m/s to the right collides with a
35-Mg car which is at rest. If after the collision the 35-Mg car is observed to
move to the right at a speed of 0.3 m/s, determine the coefficient of restitution
between the two cars.
Ans:
The momentum of the cars is conserved ,
A A B B A A B B
A
A
B A
A B
m v m v m v' m v'
20x0.5 35x0 20x v' 35x0.3
v' 0.025m / s
v' v' 0.3 ( 0.025)
e 0.65
v v 0.5 0
e 0.65
  
  
  
  
   
 


17. 17
5. A ball is dropped from an unknown height on a horizontal floor from which it
rebounds to a height of 8m. If e=0.667, calculate the height from which the
ball is dropped.
Ans:
This is case of direct central impact,
Therefore,
2
h
e
H
8
0.667
H
8
0.667
H
H 17.982m





18. 18
Thank You