This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ch means it doesn’t go as high as block 2.
Problem 3:
(a) This ﬁrst part is a collision problem in 1D, so we should use the principle of conservation of momentum. Since the blocks stick together, they have the same velocity vf
immediately after the collision:
pi = pf =⇒ m 1 v1 = ( m 1 + m 2 ) vf =⇒ vf = m1
v1 .
m1 + m2 With the numbers, this is
vf = 2. 5
2.6 m/s = 1.383 m/s ≃ 1.34 m/s.
2. 5 + 2. 2 (b) The second part of this problem can be solved using the idea of energy conservation.
The kinetic energy of the blocks is converted entirely into spring potential energy, and
This documentlost to friction. a JavaScriptenabled PDFthe collision, and
no energy is requires So considering “initial” to be right after reader
“ﬁnal” to be when the spring is maximally compressed, such as Adobe Reader. Ef = Ei =⇒ Uspr,f = Ki . The kinetic energy needs to incorporate the motion of both of the blocks, using the
value for vf that we calculated in part (a).
Uspr,f = Ki
121
kx = (m1 + m2 )v 2
2
2
m1 + m2 2
x2 =
v
k
m1 + m2
x=
v
k
Now all that’s left is the numbers:
x= 2. 5 + 2. 2
(1.383) = 0.2069 m ≃ 20.7 cm.
210 Copyright c 2012 University of Georgia. Unauthorized duplication or distribution prohibited. Physics 1111
Spring 2012 PS #8 Solutions
Page 3 of 9 Problem 4:
Linear momentum is conserved in this collision, as in any collision where external forces can
eﬀectively be neglected. However, we don’t necessarily know that kinetic energy is conserved,
so we won’t make that assumption. Since the second car is initially not moving, the initial
momentum of our twocar system is simply pi = mv .
After the collision, we can write the ﬁnal momentum as
pf = m(0.377v ) + (0.452m)vf
where vf is the unknown ﬁnal speed of the second car. Now we set pi = pf and solve for vf :
pf = pi =⇒ m(0.377v ) + (0.452m)vf = mv
vf = =⇒ 0.452mvf = mv − 0.377mv (1 − 0.377)mv
= 1.38v.
0.452m Problem 5:
We again want to use conservation of momentum in collisions as the central principle. We’re
given enough information to ﬁnd the ﬁnal momentum of the two PDF reader
This document requires a JavaScriptenabledstucktogether cars. . . but
ﬁrst we need to calculate their speed right after the collision. Sliding friction will exert a
force on Adobe
such as the cars of Reader.
ff r = µk FN = µk Mtot g.
This friction is the net force acting on the cars after the collision, so they experience a
deceleration of magnitude
Fnet
= µk g.
a=
Mtot
Knowing the distance they travel before stopping, constanta...
View
Full
Document
This note was uploaded on 08/15/2012 for the course PHYS 1111 taught by Professor Plascak during the Spring '08 term at University of Georgia Athens.
 Spring '08
 plascak
 Mass

Click to edit the document details