Two balls, A, the incident ball of mass.20 kg and Ball C, the target ball of mas
ID: 1296448 • Letter: T
Question
Two balls, A, the incident ball of mass.20 kg and Ball C, the target ball of mass .20 kg are on a horizontal table. Ball A slides at 2.3 m/s and makes a glancing collision with Ball c that is at rest at the edge of the table. The target ball, C, strikes the gloor at a point P, which is at a horizontal displacement of 0.15 m from the point of collision, and at a horizontal angle of 30 degrees from the + x axis. The tabletop is 1.2 m above the floor. What is the speed of the target ball, C, immediately after the collision Calculate the incident ball A's momentum immediately after the collision
Explanation / Answer
The time of fall from the table is
t = sqrt [ 2h / g ] = sqrt [ 2(1.2) /9.8 ] = 0.495 s
Speed is distance / time so V = 0.15/0.495 = 0.303 m/s
momentum after collision is using conservation of momentum
m1 U1 = m1 V1 + m2 V2
m1 V1 = m1 U1 - m2 V2
= (20)(2.3) - (20)(0.303)
= 39.94 kgm/s