Question

A 52.0-kg skater is traveling due east at a speed of 3.30 m/s. A 69.5-kg skater is moving due south at a speed of 7.45 m/s. They collide and hold on to each other after the collision, managing to move off at an angle ? south of east, with a speed of vf. Find the following. (a) the angle ? ° (b) the speed vf, assuming that friction can be ignored m/s

Answer 1

let east be along x axis and north be along y axis

let unit vectors along x and y axes are i and j respectively.

initial velocity of 52 kg skater=v1=3.3 i m/s

initial velocity of 69.5 kg skater =v2=-7.45j m/s

the collision is inelastic in nature

let final velocity be v3.

using conservation of momentum,

mass of first skater*velocity of first skater+mass of second skater*velocity of second skater=combined mass*velocity after collision

==>velocity after collision=(52*3.3 i +69.5*(-7.45 j))/(52+69.5)

=1.4123 i -4.2615 j

part a:

angle theta below x axis=arctan(4.2615/1.4123)=71.664 degrees

part b:

speed vf=sqrt(1.4123^2+4.2615^2)

=4.4894 m/s