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An incident ball A of mass 0.10 kg is sliding at 1.4 m/s on the horizontal tabletop of negligible friction as shown. It makes a head-on collision with a target ball B of mass 0.50kg at rest at the edge of the table. As a result of the collision, the incident ball rebounds, sliding backwards at 0.70m/s immediately after the collision.

 

a. Calculate the speed of the 0.50 kg target ball immediately after the collision. The tabletop is 1.20 m above a level, horizontal floor. The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision. 

b. Calculate the horizontal displacement 

In another experiment on the same table, the target ball B is replaced by target ball C of mass 0.10 kg. The incident ball A again slides at 1.4 m/s, as shown below left, but this time makes a glancing collision with the target ball C that is at rest at the edge of the table. The target ball C strikes the floor at point P, which is at a horizontal displacement of 0.15 m from the point of the collision, and at a horizontal angle of 30° from the +x-axis, as shown below right.

c. Calculate the speed v of the target ball C immediately after the collision. 

d. Calculate the y-component of incident ball A's momentum immediately after the collision.  

1 Answer

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(c) The time of fall is the same as before since it’s the same vertical distance. t = 0.49s 

The velocity of ball C leaving the table can be found using projectile methods. 

vx = d/t = 0.15/0.49 = 0.31m/s

(d) Looking that the y direction. 

py(before) = py(after) 

0 = pay – pcy 

0 = pay – mcvcy 

0 = pay – (0.1)(0.31)sin30 

pay = 0.015kg m/s

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