Question

A bowling ball of mass m = 1.1 kg is launched from a spring compressed by a distance d = 0.27 m at an angle of θ = 41° measured from the horizontal. It is observed that the ball reaches a maximum height of h = 3.9 m, measured from the initial position of the ball. Let the gravitational potential energy be zero at the initial height of the bowling ball 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.

Answer 1

Answer:

1) Spring constant = 2681.75 N/m

2)Velocity after launch = 13.33 m/s

Explanation:

Since the energy of system is conserved we shall equate initial energy and the energy at the top point of the ball

$E_{initial}=\frac{1}{2}kx^{2}\\\\\therefore E_{ini}=0.5\times k\times 0.27^{2}=0.03645k$

When the ball reaches it's maximum height it shall have only horizontal component of velocity while as the vertical component will become zero

Thus at the top most point the energy of the ball becomes

$E_{top}=\frac{1}{2}mv_{h}^{2}+mgh_{max}\\\\\therefore E_{top}=\frac{1}{2}m(v_{o}cos(\theta ))^{2}+mgh_{max}$

Applying values we get

$\therefore E_{top}=\frac{1}{2}\times 1.1\times (v_{o}cos(41^{o}))^{2}+1.1\times 9.81\times 3.9m$

Now since the motion of the ball is projectile motion the maximum height reached by the ball is given by

$h_{max}=\frac{v_{o}^{2}sin^{2}(\theta )}{2g}\\\\\therefore v_{o}=\sqrt{\frac{h_{max}\times 2g}{sin^{2}(\theta )}}\\\\\therefore v_{o}=\sqrt{\frac{3.9\times 2g}{sin^{2}(41)}}\\\\v_{o}=13.33m/s$

Using the velocity in the equation of energy at top we get

$E_{top}=97.75Joules$

Equating initial energy and the energy at the top we get

$0.03645k=97.75\\\\\therefore k=\frac{97.74}{0.03645}=2681.75N/m$