MIJ 18/P13/09 (S130-20 A ball of mass 0.20 kg. travelling in the x-direction at a speed of 0.5m/s. collides with a ball of mass 0.30 kg travelling in the y direction at a speed of 0.40 m/s The two balls stick together after the collision, travelling at ap angle to the x-direction 10.30kg 10 40m ) direction 0.50 ms D mum 6205 do 0.20 kg y direction barco 0-12 What is the value of 6? A 39 B 40° C 50 D 51
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Explanation:
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. momentum impulse jee jee mains Share It On 1 Answer +1 vote answered May 27, 2019 by AmreshRoy (69.5k points) selected May 27, 2019 by Vikash Kumar (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/sRead more on Sarthaks.com - https://www.sarthaks.com/378167/an-incident-ball-mass-sliding-on-the-horizontal-tabletop-of-negligible-friction-as-shown