Ball B will have to spend 1.5 seconds going up and 1.5 seconds coming down, since Ball A takes three seconds to go down. (Up and down times are equal)
Ball A's average velocity while going down is 6 m/s. Its final velocity is therefore 12 m/s, since it accelerates uniformly.
Let the distance Ball B travels up the ramp be x*18 m. (x is a dimensionless fraction)
Final velocity of Ball B:
VBmax = (18x/1.5)*2 = 24 x
Also,since the achieved maximum velocity is proportional to sqrtx,
VBmax = sqrt x*VAmax = sqrtx*12 m/s
24x = sqrtx*12
sqrtx = (1/2)
x = 1/4
Ball B travels 1/4 of the way up the ramp.
For the projection velocity of ball B,
VBmax = [sqrt(1/4)]*12 = 6 m/s.
A ball rolling on an inclined plane dors so with constant acceleration. One ball, A, is released from rest on the top of an inclined plane 18m long to reach the bottom 3s later. At the same time as A is released, another ball, B, is projected up the same plane with a certain velocity. B is to travel part way up the plane, stop, and return to the bottom so as to arrive simultaneously with A. calculate the velocity of projection of B.
1 answer
1 answer