16. A 2-kg box is released from rest from the top of a 5-m-high frictionless ramp. At the bottom of the ramp the box has an elastic collision with an 8-kg box that is initially at rest. The 2-kg box bounces off the 8-kg box and goes back up the ramp. To what height does the 2-kg block return up the ramp before coming to rest? A. 0.4 m B. 0.8 m C. 1.0 m D. 1.8 m E. 2.5 m
16. A 2-kg box is released from rest from the top of a 5-m-high frictionless ramp. At the bottom of the ramp the box has an elastic collision with an 8-kg box that is initially at rest. The 2-kg box bounces off the 8-kg box and goes back up the ramp. To what height does the 2-kg block return up the ramp before coming to rest? A. 0.4 m B. 0.8 m C. 1.0 m D. 1.8 m E. 2.5 m
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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16 please with provided formula sheet easy way to explain please
Transcribed Image Text:16. A 2-kg box is released from rest from the top of a 5-m-high frictionless ramp. At the bottom of the ramp the box has
an elastic collision with an 8-kg box that is initially at rest. The 2-kg box bounces off the 8-kg box and goes back up the
ramp. To what height does the 2-kg block return up the ramp before coming to rest?
A. 0.4 m
ving
B. 0.8 m
C. 1.0 m
D. 1.8 m
E. 2.5 m
ease
ick t
101
objec
above a
ring, w
6.
E) 960
Transcribed Image Text:ne exam is
v2
Circular motion: a. =
R
Enet = ma;
Weight: Fg = mg, ;
W
Pavr
At
g = 9.8 m/s²;
1
Kinetic energy: K =
-m v²;
Potential energy: Ug=mgy
E = Er
E = U+K
Rotational motion:
0 = =
1 rev = 2n rad;
v = o r;
at = a r;
a, =
0 = on t+at?;
may b<
T= rFsino;
0 = 00 + at;
2θα-ω?-0
T=rxF;
Iş mR2
Στ- Ια;
Ipointmass = mr?
Idisk =
mR2
Ipipe =
1
Ihoop = mR?
Irod(center) =
mL?
Irod(end) =
mL?
Iball = =mR?
5
Ishell = =mR?
12
3
dW
P =
dt
I = Icom + MD?
work: W=t 0;
2
W =
Pavr =
At
2
1
-m Vcom 2
Rolling:
K =
2
T= f,R
F,max = H,Fn
Vcom = Ro
Incline: Fe=mgsin0 F=mgcøs0
Angular momentum:
Lpoint mass =m rxv
L= mrvsin 0;
L=m (r,vy - ryVz)k
L= Io
m,X1 +m2X2
miy1+m2y2
L = Lf
I 01 =I 202
X com =
Y com =
m1 +m2
m1 +m2
1
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