Faculty of
Engineering
and
Department of
Physics
ENPH 131
Last Name:
Seminar 9
Linear Impulse and
Momentum
First Name:
ID Number:
Starting March 24, 2014
Seminar Section:
Please do all work on handout and box final numerical answers. Total=20 marks.
1. [5 Points]
Answer the following questions by selecting one of the choices provided. Circle your
choice.
a) Two iceboats (shown below) race on a frictionless horizontal lake. Boat A and boat B
have masses of m and 2m, respectively. Each iceboat has an identical sail, so the wind

exerts the same constant force, F on each iceboat. The two boats start from rest and
cross the finish line a distance, s away. Which boat crosses the finish line with a
momentum of greater magnitude?
Boat A
Boat B
Both have same momentum
Insufficient information provided
b) A soft tennis ball and a hard golf ball are thrown against a rigid wall. Curves A and B of
impulse force versus time were prepared for each ball during their impact. The curves
show a rise in the force, up to a maximum value during impact of the balls on the wall.
Circle the correct answer below.
Curve A = tennis ball
Curve B = golf ball
Curve A = golf ball
Curve B = tennis ball
Insufficient information provided
c) From part b), what is the area under the curves?
Momentum
Kinetic Energy
Impulse
Insufficient information provided
Solution [2 Points each for parts a) and b); 1 Point for part c)]
a) Two iceboats (shown below) race on a frictionless horizontal lake. Boat A and boat B
have masses of m and 2m, respectively. Each iceboat has an identical sail, so the wind

exerts the same constant force, F on each iceboat. The two boats start from rest and
cross the finish line a distance, s away. Which boat crosses the finish line with a
momentum of greater magnitude?
1
Boat A
Boat B
Both have same momentum
Insufficient information provided

The principle of linear impulse and momentum states that mv1 
t2



Fdt  mv 2 . When the
t1
magnitude of the impulsive force is very large, compared to other forces acting on the



body, the principle of linear impulse and momentum becomes mv1  F t 2  t1   mv 2 .


Since the boat started from rest ( mv1  0 ), the momentum ( p ) at the finish line is defined




as p  mv 2  Ft . Both iceboats experience the same average applied force ( F ), but the
more massive boat of mass, 2m, takes a longer time to accelerate, and ultimately, a longer
time to cross the finish line. Therefore, Boat B will cross the finish line with greater
momentum.
b) A soft tennis ball and a hard golf ball are thrown against a rigid wall. Curves A and B of
impulse force versus time were prepared for each ball during their impact. The curves
show a rise in the force, up to a maximum value during impact of the balls on the wall.
Circle the correct answer below.
Curve A = tennis ball
Curve B = golf ball
Curve A = golf ball
Curve B = tennis
Insufficient information provided
The rigid, hard golf ball will have a large maximum force and the impact will last a short
amount of time. Curve A represents this situation. The soft tennis ball will have a lower
maximum force and the impact time will be longer. Curve B represents this situation.
c) From part b), what is the area under the curves?
Momentum

Impulse is J 
Kinetic Energy
t2

Impulse
Insufficient information provided

Fdt = area under F versus t curve.
t1
2
2. [7 Points]
A coach’s bat exerts a horizontal force on a 0.145-kg baseball of


F t   1.60 x 10 7 N/s t  6.00 x 10 9 N/s 2 t 2 i ,

 
 
between t = 0 and t = 2.50 ms.
At t = 0, the baseball’s velocity is



v  40.0i  5.0 j  m/s.
Determine: a) the impulse exerted by the bat on the ball during the 2.50 ms that they are in
contact,
b) the average force exerted by the bat on the ball during this time interval, and,
c) the velocity of the baseball at t = 2.50 ms.
(Hint: Magnitudes and directions should be specified)
Solution
a)
The principle of linear momentum and impulse is
 

mv1  J  mv 2

mv1 
or
t2



Fdt  mv 2 .
t1

The impulse, J is

J
t2


Fdt
[1 Point]
t1

J
t2  2.50 ms
 1.60 x 10
7
 
 

N/s t  6.00 x 10 9 N/s 2 t 2 i dt
t1 0




t  2.50 ms
3 2
 
t2
7
9
2 t 
J   1.60 x 10 N/s
 6.00 x 10 N/s

2
3 0


i
2
3
3
 

2.50 ms 2  1 s 
 1s  
7
9
2 2.50 ms 
x
x
J   1.60 x 10 N/s
  6.00 x 10 N/s
 i
2
3
 1000 ms 
 1000 ms  




J  18.8 N - s i  18.8 kg - m/s i
[2 Points]

b)



For time-dependent functions, the time-weighted average force is defined as

1
Fave 
t

Fave 
t2


Fdt 
t1
t2

1
Fdt
t 2  t1 

[1 Point]
t1

J
t 2  t1 



1000 ms
18.8 kg - m/s i
Fave 
x
2.50 ms  0 ms
1s



Fave  7.50 x 10 3 kg - m/s 2 i  7.50 x 10 3 N i

 

[1 Point]
3
c)
From the principle of linear momentum and impulse, in general,
 

mv1  J  mv 2 .


The fact that F t  is defined by the i unit vector suggests that the force is exerted only in the x-direction on
the ball. Therefore, only the x-component of momentum and the x-component of velocity will change. So,



mv1, x  J x  mv 2, x

Jx


v 2, x  v1, x 
m




18.8 kg - m/s i
v 2, x  40.0i m/s 
0.145 kg




[1 Point]
v 2, x  40.0i m/s  129.7i m/s  89.7i m/s .
So, the velocity of the ball is



v 2  89.7i  5.0 j  m/s
[1 Point]
Note that the y-component of the velocity remained unchanged.
4
3. [8 Points]
The 40-kg block shown in the schematic below is initially moving upward at a speed of 2.5
m/s. The block is connected to a 15-kg and a 10-kg block by way of a massless cord that is
routed over weightless, frictionless pulleys. All the blocks are in contact with a solid
surface, and the coefficient of friction between the blocks and the surfaces are 0.1.
a) Draw the free body diagrams for the blocks.
b) What constant value of the applied force, P will produce an upward velocity of 5 m/s in 12
s for the 40-kg block (Block A)?
(Hint: The tension forces in the cord that connects Blocks A and B are different from those
that are in the cord that connects Blocks B and C)
Solution
The application of forces on a body that produce a change in velocity of the mass over a specified
time period may imply that this is a problem of linear impulse and momentum. The principle of
linear impulse and momentum is

mv1 
t2



Fdt  mv 2 .
t1
In this problem, multiple forces act on the blocks and they all need to be considered. Consider also
the impulse and momenta of the blocks along the line of travel. Drawing a free-body diagram of the
forces acting on the blocks may be helpful to visualize the problem. Take the positive direction for
each block as being in the direction of motion. So,
[3 Points]
(1 Point for each block)
The tension forces in the cord cancel. Also, there is no friction force acting on Block A because there
is no normal force. Then,
t2




 
Fdt  m total v 2  m total v1  m total v 2  v1 
t1
5
 m
Ag

 
 f B N B  f C N C  mC g sin 45 o  P t  m total v 2  v1 
[2 Points]
Note that the component of the weight of Block C in the direction of motion and the friction force on
Block C is
W x  mC g sin 45 o  69.4 N
F f  f C mC g cos 45 o  6.94 N
 m
Ag

 
 f B m B g  f C mC g cos 45 o  mC g sin 45 o  P t  m A  m B  mC v 2  v1 
P
P
[2 Points]
m A  m B  mC v2  v1 
t

 m A g  f B m B g  f C mC g cos 45 o  mC g sin 45 o

40 kg  15 kg  10 kg 5 m/s  2.5 m/s
12 s
 40 kg  9.81 m/s  0.115 kg  9.81 m/s 2  0.110 kg  9.81 m/s 2 cos 45 o  10 kg  9.81 m/s 2 sin 45 o


2



P = 358 N





[1 Point]
6

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