Question

An adventurer, Jason, with a mass of 65 kg snowboards down a slope, starting at rest from a height of 10m. When he arrives at the bottom of the slope, where it is level, he gives a push to his best friend, Jack who also weighs 65 kg. Assume there is no friction on the snow and that the impact between the friends is nearly elastic:

1. What is the total energy of the system at the beginning when Jason is at the top of the mountain and Jack is at the bottom, not moving?

2. What is Jason's speed when he reaches the bottom?

3. What is Jason's momentum at the bottom when the speed is V(bottom)

4. What is the speed of Jack after they collide?

Answer 1

This problem is based on conservation of energy and is explained below..
mass of Jason mis = 65kg Height of Slopediy= lom. mas of Jack&may = 65 kg . → when Joson Snowboards down a slope from a height of lom then this gravitational potential energy Converts to Kinetic energy. Now when he Collides with Jack then by cons of momentum both will gain some momentum as explained. Initially. Initially Halom H=lom I). Initially when Joson is at top and Jack is at bottom then their Kinetic energy is zero. but. as Joson is above the reference, he has gravitation ronal Potential energy and gravitational P. E of Jack {mat is zero. Total energy of System - migh (6500 I
k
21 By Conservation of Energy. kili + P. Ei = kept P. Ef OxgH = $ $2 to vil ag H "= fagh = 52(10(10) =/loja mp- 3). Momentum of Joson's family is of bil = 365 kg {{ lojal I k, = 65.52 kg m/s. 4). As Collision blue Jack Amy & Jason&miy is elostic. Е, Соn, ѕg mоmе аn y = 3 у — ""}. Als e= 3-v2 - Coeff. of restititution. for elastic Collision è=1, "="3-02 @ Solving eq o 22. 20,- 203 → NAV . if vi= uz then vazo. - so speed of Jack 2m2f after Collision =Y=lloran