Problem 1: A bowling ball of mass m = 1.9 kg is launched from a spring compresse
ID: 1518355 • Letter: P
Question
Problem 1: A bowling ball of mass m = 1.9 kg is launched from a spring compressed by a distance d = 0.22 m at an angle of = 38° measured from the horizontal. It is observed that the ball reaches a maximum height of h = 3.5 m, measured from the initial position of the ball. Let the gravitational potential energy be zero at the initial height of the bowling ball.
Randomized Variables m = 1.9 kg d = 0.22 m = 38° h = 3.5 m
Part (a) What is the spring constant k, in newtons per meter?
Part (b) Calculate the speed of the ball, vo in m/s, just after the launch.
Explanation / Answer
height of the ball when it leaves the spring, h = d*sin(38)
= 0.22*sin(38)
= 0.135 m
Hmax = 3.5 m/s
let vo is the velocity of the ball when it leaves the spring.
voy = vo*sin(38)
vox = vo*cos(38)
Hmax - h = voy^2/(2*g)
voy = sqrt(2*g(Hmax - h))
= sqrt(2*9.8*(3.5 - 0.135))
= 8.12 m/s
so, vo = voy/sin(38)
= 8.12/sin(38)
= 13.2 m/s <<<<<<<----------------Answer for part b)
a) Apply conservation of energy
0.5*k*d^2 = m*g*h + (1/2)*m*v^2
k = (m*g*h + (1/2)*m*v^2)/(0.5*d^2)
= (1.9*9.8*0.135 + (1/2)*1.9*13.2^2)/(0.5*0.22^2)
= 6944 N/m <<<<<<<----------------Answer for part a)