Answer to Question #287718 in Mechanics | Relativity for Teejay

Question

Two identical balls, A and B, are released from rest from the same height, 2.8 m, along two smooth inclined planes that touch the horizontal ground. Ball A is released on a plane inclined at 30° to the horizontal. The plane on which ball B is released is inclined at 60° to the horizontal. Ignore energy due to rotation of the balls. (a) What is the speed of each ball at the bottom of their respective inclined planes? (b) Which ball would reach the bottom first? Explain your answer.

Answer 1

a) The speed is the same for both balls:

$$v=\sqrt{2gh}=\sqrt{2×9.8×2.8}=7.4 m/s$$

b) The first ball to reach bottom will be the ball released along a 60°-slope because it is more vertical, hence, the acceleration due to gravity will influence its motion more.