Collision problem?

Question

A 65.0 kg person, running horizontally with a velocity of +3.68 m/s, jumps onto a 12.2 kg sled that is initially at rest.
(a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away in m/s
(b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?

Can someone help me work this?

Answer 1

For part a, use conservation of momentum to solve:

m_1*v_1 = m_2*v_2 where,

m_1 = mass of person
v_1 = velocity of person
m_2 = total mass of person and sled
v_2 = velcoity of m_2

We are solving for v_2, thus:

v_2 = m_1*v_1/m*2 --------> (65 kg)(3.68 m/s)/(65+12.2 kg)

Solving you get, v_2 = 3.1 m/s

-------------

For part b, you can use conservation of energy to solve. When the sled and person come together, they have a given velocity. There is initial kinetic energy which is then dissipated due to friction.

The setup is as follows:

(1/2)m*v^2 = (Frictional force)(distance) = u*N*d where N = mg

Thus, (1/2)*m*v^2 = u*m*g*d. Using g = 9.8 m/s^2, we have all the components of this equation. Now solve for coefficient of friction, u.

u = (m*v^2)/(2*m*g*d) = v^2/(2*g*d) = (3.1)^2/(2*9.8*30)

Solving you get u = 0.016

-----------------------

Hope this helps

Answer 2

Hints
For a, use conservation of momentum. m1v1 = (m1+m2)V2
For b, find the acceleration from d=vt+at^2/2. The frictional force is F=ma = uN, where N is the normal force and u is the desired coefficent of friction.