The 2-kg ball A is thrown so that it travels horizontally at 10 m/s when it strikes the 6-kg block B as it is traveling down the incline at 1 m/s. If the coefficient of restitution between the ball and the block is e=0.6, determine the speeds of the ball and the block just after impact.

answer: i pick a

explanation:

iron would move it is magnetic there is a mag field around the current in the wire

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft