Question

As shown in the figure below, object m1 = 1.50 kg starts at an initial height h1i = 0.276 m and speed v1i = 4.00 m/s, swings downward and strikes (in an elastic collision) object m2 = 4.70 kg which is initially at rest. Determine the following.

Answer 1

Solution Write the expression for initial energy of ball. 2 2mgn 0.5m vo- 2(9.81) (0.276)+4* -4.6276m/s For elastic collision, write the velocities post collision of any two bodies m+m m+m Foru,-0, the body 2 is at rest Mite the expression for post collision velocities

(m+m2) 4.6276(1.5-4.7) --2.388 m/s 2mui (m+m2) 2(1.5) 4.6276) - 2.239m/s Hence, the ball 1 goes in the negative x direction and ball 2 goes in the positive x direction. c) Determine the height to which each ball swings is obtained through energy conservation. mgh-0.5(m) 0.5(1.5)(2.388) 4.2769 0.29m 0.5(1.6)(2.239) 4.010 - 0.255m