Copy of NOTA TUTORIAL SP015 CHAPTER 3 20_21

TOPIC 3: MOMENTUM AND IMPULSE Impulse, J also related with momentum, p as stated in;
3.1 Momentum and Impulse
3.2 Conservation of Linear Momentum            
J p pf pi mv mu
3.1 Momentum and Impulse
Learning Outcome: Example 3.1.1:

a) Define momentum and impulse,
b) Solve problem related to impulse and impulse-

momentum theorem.
c) Use F-t graph to determine impulse.

Momentum

 Defined as the product of a mass, m and the

velocity, v.  
p mv


 The direction of the momentum is same as the Figure 3.2 shows that a 0.20 kg tennis ball strikes the
direction of the velocity. wall horizontally with a speed of 100 m s1 and it bounces
off with a speed of 70 m s1 in the opposite direction.
 The S.I. unit of linear momentum is kg m s-1 or N s.
a. Calculate the initial and final momentum of the ball.
Impulse
b. Calculate the impulse delivered to the ball by the all,
 Defined as the product of force, F and the time
c. If the ball is in contact with the wall for 10 ms,
interval, Δt.  determine the impulsive force exerted by the wall on the
ball.
J  Ft
Solution:
 Impulse is a vector quantity and its direction is same
as the direction of force on the object. a)

 The S.I. unit of impulse is N s or kg m s1.
 Impulse, J can be determined from the graph of

impulsive force, F against time, t as shown in Figure
3.1.

Where: b) From the impulse-momentum theorem:

J  p  p f  pi
 (14)  (20)
 34 N s

c) Given the contact time: 3.2 Conservation of linear momentum

t  10 103 s Principle of conservation of linear momentum
J  Ft
“In an isolated (closed) system, the total
 34  F (10 103) momentum of that system is constant.”
F  3400 N
“When the net external force on a system is
(since u is positive, negative sign of J and F value means, zero, the total momentum of that system is
the direction of impulse and impulsive force is opposite constant.”
to the initial direction of the tennis ball)
OR
Example 3.1.2:
 According to the principle of conservation of linear
An estimated force-time curve for a tennis ball of mass momentum, we obtain that total momentum of the
60.0 g struck by a racket is shown in Figure 3.3. system is constant
Determine:
a. the impulse delivered to the ball, The total of initial = the total of final
b. the velocity of the ball after being struck, assuming the
ball is being served so it is nearly at rest initially. momentum   momentum
pi pf
Solution: 

m  60.0 103 kg If  F  0

a. From the force-time graph Or
In a closed system
J  area under the F  t graph
Conditions for Elastic and Inelastic Collisions
 J  1 1.8  0.2103 18 103
2 TYPE OF CONDITIONS
J  14.4 N s COLLISION

b. Given the ball’s initial speed, ELASTIC Total momentum is conserved.
Total kinetic energy is conserved.
u0
J  p  p f  pi Total momentum is conserved.
 mv  mu
 m(v  u) INELASTIC Total kinetic energy is not
conserved.
 14.4  60.0 103 v  0
 Use the condition of total kinetic energy to
v  240 m s1 determine/prove the type of collision.

MISCONCEPTION ABOUT TYPE OF a)
COLLISION:
mAuA  mBuB  mAvA  mBvB
Elastic collision:
25010.6 260-10.6  250(8.4)  260vB
 The collision is elastic because after the collision,
the objects move separately. vB  7.7 m s1 (to the right)

Inelastic collision: b) The type of the collision can be identified by
calculating the initial and final total kinetic energy of the
 The collision is inelastic because after the collision, system.

the objects stick/move together.  K.Ei  1 mAuA2  1 mBuB 2  28651 .8 J
2 2
IN FACT:
 K.Ef  1 mAvA2  1 mBvB 2 16527.7 J
Elastic collision: 2 2

 The collision is elastic only when the total kinetic Since:
energy before collision equal to total kinetic energy
after collision.  K.Ei   K.Ef

 The objects move separately after collision. the type of the collision is INELASTIC.

Inelastic collision: Note:

 The collision is inelastic only when the total kinetic Final condition of two bodies that experience inelastic
energy before collision not equal to total kinetic collision is not necessary to be sticking together.
energy after collision.
Only for completely (or perfectly) inelastic collision, the
 Total kinetic energy after collision < total kinetic bodies will finally stick together where final velocity for
energy before collision. both of them are same.

 The objects move separately (or stick together after Linear Momentum in Two Dimension Collision
collision – perfectly inelastic collision). (from the top view)

Linear Momentum in One Dimension Collision Example 3.2.2:

Example 3.2.1:

Figure 3.5 Figure 3.6

Figure 3.5 shows car A and car B are moving directly to A tennis ball of mass 250 g moving with initial velocity
each other at a same speed of 10.6 m s-1 and finally they 20 m s1 collides with a soccer ball of mass 900 g initially
will collide. Right after the collision, car A bounces at rest. After the collision, the tennis ball is deflected 50
backward at a speed of 8.4 m s-1. Given that the mass of from its initial direction with a velocity 4 m s1 as shown
car A and car B are 250.0 kg and 260.0 kg respectively. in Figure 3.6. Determine and sketch the velocity of soccer
ball after the collision.
a) Calculate the final velocity of car B.
Solution:
b) What is the type of the collision?
m1  0.250 kg; m2  0.900 kg; u1  20 m s1;
Solution:
u2  0; v1  4 m s1; θ1  50
Given;
From the principle of conservation of linear momentum:
mA  250 kg; mB  260 kg;
uA  10.6 m s1;uA  10.6 m s1
vA  8.4 m s1

    
pi pf

The x-component of linear momentum:

   
pix p fx

m1u1x  m2u2x  m1v1x  m2v2x

0.25020 0  0.250v1 cosθ1 0.900v2x

 5  0.250 4cos50  0.900v2x

v2x  4.84 m s1

The y-component of linear momentum:

   
piy p fy

0  m1v1y  m2v2y

 0  0.250  4sin 50  0.900v2y

v2 y  0.851 m s1

Magnitude of the soccer ball:

  v2  v2x 2  v2y 2

v2  4.842  0.8512  4.91 m s1

Direction of the soccer ball:

θ2  tan 1 v2 y   tan 1 0.851 
v2 x  4.84 

θ2  9.97 from positive x-axis
anticlockwise

The sketch of the final velocity of the soccer ball:

The velocity of the soccer ball after the collision is 4.91
ms-1 with the angle θ= 9.97° as shown above.