Question

A small ball of clay of mass m hangs from a string of length L (the other end of which is fixed). A seond ball of clay of mass m/3 is to be launched horizontally out of a spring with spring constant k. Once launched, the second ball will collide with and stick to the hanging ball, and they'll follow a circular path around the fixed end of the string.

A) Determine an expression for the distance (change in x) that the spring should be compressed by, in terms of g and th define variables, that will allow for the tension in the string to be zero at the top of the circular motion.

Answer 1

when tension becomes zero, minimum speed at thr top of the loop,

v_top = sqrt(g*L)

minimum speed at the bottom of the loop, v_bottom = ssqrt(5*g*L)

let u is the speed of second ball just before the collision.

Apply conservation of momentum

(m/3)*u = (m/3 + m)*v_bottom

(m/3)*u = (m/3 + m)*sqrt(5*g*L)

u = 4*sqrt(5*g*L)

let x is the comression of the spring and k is the spring constant.

now apply conservation of energy

initial elastic potential energy of the spring = kinetic energy gained by the second ball

(1/2)*k*x^2 = (1/2)*m*u^2

(1/2)*k*x^2 = (1/2)*m*(4*sqrt(5*g*L))^2

k*x^2 = 4*m*5*g*L

x = sqrt(20*m*g*L/k) <<<<<<<------Answer