A curling stone with a mass of 7.00 kg moves across the ice at a speed of 1.35 m
ID: 2201819 • Letter: A
Question
A curling stone with a mass of 7.00 kg moves across the ice at a speed of 1.35 m/s in the positive x direction. It collides with another curling stone with the same mass. Stone 2 has a 0 m/s initial velocity. After the collision stone 1 has a speed of 0.457 m/s at an angle of 54 degrees to the left of the x axis. a) What is the initial kinetic energy? b) What is the initial momentum? (magnitude and direction) c) What is the speed and direction of stone 2 after the collision? d) What is the total momentum perpendicular to the x axis after the collision?Explanation / Answer
x component: V1i=V1f cos?+ V2f cos? 1.7m/s=0.800m/s61.9 + V2f cos61.9 V2fcos? = -1.7m/s+0.800m/s cos61.9 V2fcos?= -1.32m/s V2fcos?= 1.32m/s [backwards] y- component 0+0=mv1f sin ?-m2v2fsin ? m2v2fsin ?= (0.800m/s) sin 61.9* v2fsin?=(0.800m/s) sin 61.9* v2fsin?=0.7057m/s (-1.32m/s)^2 + (0.7057m/s)^2 = 1.7424m/s +0.4980m/s=2.24041249m/s =2.24041249m/s ?tan =0.7057/-1.32 ?=tan^-1= 28*