Conservation of momentum and energy

[JJBrian]

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?

 

[JJBrian]

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

 

[tomcue]

a) Conservation of momentum and energy tells us that in a closed system, the total momentum and energy remain constant. In this scenario, the cart is the only object in motion and there are no external forces acting on it, so it can be considered a closed system.

The acceleration due to friction can be calculated using the equation a = (v2-v1)/t, where v2 is the final velocity (0.4 m/s), v1 is the initial velocity (0.5 m/s), and t is the time interval (3 seconds). This gives us an acceleration of -0.033 m/s^2.

The frictional force can be calculated using the equation F = ma, where m is the mass of the cart (0.5 kg) and a is the acceleration calculated above. This gives us a frictional force of -0.0165 N.

The work done by friction can be calculated using the equation W = F*d, where F is the frictional force calculated above and d is the displacement of the cart (1.4 m - 0.5 m = 0.9 m). This gives us a work done by friction of -0.01485 J.

The change in kinetic energy of the cart can be calculated using the equation ΔKE = 1/2*m*(v2^2-v1^2), where m is the mass of the cart (0.5 kg) and v2 and v1 are the final and initial velocities, respectively. This gives us a change in kinetic energy of -0.025 J.

b) In an elastic collision, both momentum and kinetic energy are conserved. Using the equations for conservation of momentum and kinetic energy, we can solve for the final velocity of the first cart.

Conservation of momentum: m1v1 + m2v2 = m1v1' + m2v2', where m1 and m2 are the masses of the first and second cart, respectively, and v1, v2, v1', and v2' are the initial and final velocities of each cart.

Conservation of kinetic energy: 1/2*m1*v1^2 + 1/2*m2*v2^2 = 1/2*m1*v1'^2 + 1/2*m2*v2'^2

Plugging in the given values and solving the system of equations, we get a final velocity of 0