A ball of mass m = 1.50 kg is released from rest at a height h = 73.0 cm above a

Question

A ball of mass m = 1.50 kg is released from rest at a height h = 73.0 cm above a light vertical spring of force constant k as in Figure [a] shown below. The ball strikes the top of the spring and compresses it a distance d = 7.80 cm as in Figure [b] shown below. Neglecting any energy losses during the collision, find the following. in (a) Find the speed of the ball just as it touches the spring. 3.783 m/s (b) Find the force constant of the spring. 4400 Your response differs from the correct answer by more than 10%. Double check your calculations. N/m Need Help? Chat About It Submit Answer Save Progress Practice Another Version

Explanation / Answer

let the initial height (the relaxed height of the spring) of the spring be our referrence level of zero potential energy.

as there is no energy loss during collision,

total energy of the system remains constant at all the points.

total energy consists of 3 parts:

1)potential energy of the mass m

2)kinetic energy of mass m

3) potential energy of the spring

at the initial point(when the ball is at height h)

speed of the ball=0==>kinetic energy of the ball=0

spring is relaxed==>potential energy of the spring=0

hence total energy of the system=potential energy of the ball

=m*g*h

=1.5*9.8*0.73

=10.731 J

part a:

consider the moment when the ball just touches the spring

height of the ball=0==>potential energy of the ball=0

compression of the spring=0==>potential energy of the spring =0

hence total energy of the system=0.5*1.5*v^2 where v is the speed of the ball

as energy remains constant, 0.75*v^2=10.731

==>v=sqrt(10.731/0.75)=3.7826 m/s

part b:

consider the moment when spring is compressed by 7.8 cm

then potential energy of the ball=-m*g*0.078 (as spring's relaxed position is our referrence for zero potential energy)

=-1.5*9.8*0.078

=-1.1466 J

potential energy of the spring=0.5*k*0.078^2=3.042*10^(-3)*k J

speed of the ball=0==>kinetic energy of the ball=0

hence total energy of the system=-1.1466+3.042*10^(-3)*k

as energy remains constant,

-1.1466+3.042*10^(-3)*k=10.731

==>3.042*10^(-3)*k=11.878

==>k=3904.5 N/m

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