- #1
JJBrian
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Homework Statement
a) At t=1 second, a 0.5 kg cart is at x=0.5 m, traveling at 0.5 m/s. At t=3 seconds, the cart is
at 1.4 m, traveling at 0.4 m/s. What is the acceleration due to friction, the frictional force and
the work done by friction? What is the change in kinetic energy of the cart?
b) If a cart of mass 0.6 kg, traveling a 0.5 m/s collides elastically with a cart of mass 0.4 kg,
initially at rest, what is the final velocity of the first cart?
c) If the two carts above stick together after the collision, what is the final velocity of the first
cart?
d) What is the final velocity of the first cart in (b) if the second cart has mass 1.5 kg?
e) What is the final velocity of the first cart in (c) if the second cart has mass 1.5 kg?
Homework Equations
Elastic collison formula
Inelastic collision
Impulse
W=-FD
Change in kinetic energy formula
The Attempt at a Solution
ok here is my attempt. I need someone to check if I am using the right equation for each problem.
part a)
i used vf^2-vi^2 = 2ad
to get an acceleration of .05
To find the firctional force i used
F = -ma F= (.5)(.05)
Ff = -.025N
work done by friction
W = -Fd
W =-(-.025)(.05)
W = 0.0125J
change in ke
Ke =1/2mvf^2-1/2mvi^2
KE=1/2(.5)(0.5)^2-.5(0.5)(0.4)^2
Kef = .0225
part b)
v1f = vi*(m1-m2)/(m1+m2)
0.5*(0.6-0.4)/(0.6+0.4)
v1f=0.1m/s
part c)
v1f = vi*(m1)/(m1+m2)
(0.5)(.6)/(0.6+0.4)
v1f=0.3m/s
part d
v1f = vi*(m1-m2)/(m1+m2)
0.5*(0.6-1.5)/(0.6+1.5)
v1f=0.2143m/s
part e
v1f = vi*(m1)/(m1+m2)
(0.5)(.6)/(0.6+1.5)
v1f=0.143m/s