home
***
CD-ROM
|
disk
|
FTP
|
other
***
search
/
Investigating Forces & Motion
/
Investigating Forces and Motion (1998)(Granada Learning).iso
/
data
/
topic7
/
example.dat
< prev
next >
Wrap
INI File
|
1998-02-16
|
2KB
|
59 lines
[general]
[page1]
type:3
caption:\
A baseball with a mass of 0.1 kg is pitched at 30 m/s. The batsman \
hits it straight back towards the pitcher at 40 m/s. What are the \
ball's initial momentum, final momentum, and change in momentum ? \
Taking direction towards the bat as positive, drag the correct values \
into place.<p>
feedback:\
Well done.<p>\
Momentum = mass x velocity<p>\
Initial momentum = 0.1 x 30 = 3.0 kg m/s<p>\
Final momentum = 0.1 x -40 = -4.0 kg m/s<p>\
Change in momentum = final momentum - initial momentum<p>\
<center>= -4.0 - 3.0</center><p>\
<center>= - 7.0 kg m/s</center><p>
source:7ex1a, 7ex1b, 7ex1c
target:7ex1a1, 7ex1b1, 7ex1c1
[page2]
type:0
caption:\
<img src="7g12" align=center><p>\
When a baseball is struck its momentum changes by 7.0 kg m/s. If the \
ball is in contact with the bat for 0.005 s what is the average force \
of the bat on the ball?<p>\
Impulse = <I>Ft</I> = change in momentum<p>\
Thus <I>F</I> = change in momentum/<I>t</I><p>\
= 7.0/0.005<p>\
= <U>1 400 N.<p>\
</U><p>
[page3]
type:0
caption:\
<img src="7g13" align=center><p>\
Two trucks travelling on the same smooth track collide and link \
together. If the masses of the trucks are 2.0 kg and 5.0 kg and their \
velocities are as shown in the diagram, what is their combined \
velocity after the collision?<p>\
Using the principle of the conservation of momentum<p>\
<I>m</I><I><SUB>1</SUB></I><I>u</I><I><SUB>1</SUB></I> + \
<I>m</I><I><SUB>2</SUB></I><I>u</I><I><SUB>2</SUB></I> = \
<I>m</I><I><SUB>1</SUB></I><I>v</I><I><SUB>1</SUB></I> + \
<I>m</I><I><SUB>2</SUB></I><I>v</I><I><SUB>2<p>\
</SUB></I><p>\
2.0 x 5.0 + 5.0 x (-3.0) = 2.0<I>v<SUB>1</SUB></I> + \
5.0<I>v<SUB>2</SUB></I><p>\
Now the trucks link together so <I>v<SUB>1</SUB></I> = \
<I>v<SUB>2</SUB></I> = <I>v </I><p>\
Thus<p>\
10 - 15 = 7.0<I>v<p>\
</I><p>\
and <I>v</I> = -5.0/7.0 m/s<p>\
= <U>-0.71 m/s.</U><p>