A 10 pound bowling ball with a speed of 2.0 meters/second hits a 6 pound bowling ball. If the 10 pound bowling ball stops after the collision, what is the resulting speed of the 2nd ball in meters/second
use conservation of momentum to answer this.
--> (m1*v1 of 10lb ball)+(m2*v1 of 6lb ball) = (m1*v2 of 10lb)+(m2*v2 of 6lb)
--> 10lb=4.5kg; 6lb=2.7kg
--> (4.5kg *2m/s)+(2.7kg*0m/s) = (4.5kg*0m/s) + (2.7kg*v2)
-->9+0 = 0+2.7*v2
v2 = 3.33m/s