Homework Help: Conservation of momentum and energy

1. Oct 19, 2009

JJBrian

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Elastic collison formula
Inelastic collision
Impulse
W=-FD
Change in kinetic energy formula

3. The attempt at a solution
ok here is my attempt. I need someone to check if Im using the right equation for each problem.

part a)
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