Calculating Projectile Velocity with Spring Compression

[BlueDevil14]

I have attempted to solve this problem repeatedly, but I cannot get a correct answer. Any help would be wonderful.

The Problem
A certain kind of projectile launcher has a spring constant of 50 N/m and you compress the spring 18 cm before it locks. When compressed, the spring is 2 cm tall. The piston and spring each weigh 50 g and the ball weighs 10 g. If you launch the ball vertically, what do you expect the velocity of the ball to be as it leaves the launcher? Don't forget the gravitational potential energy gained as each object's center of mass moves vertically.

The correct answer is 4.13 m/s

My attempt at a solution

Potential energy at full compression must equal kinetic energy minus gravitational potential energy at the point when the ball is released

At full compression: U = 1/2*k*d^2 - m*g*h1 where d is displacement
At release: K=.5*M*v^2 where M is effective mass (.077 kg)
U=-m*g*h2

Therefore: 1/2*k*d^2 - m*g*h1=.5*M*v^2-m*g*h2
or 1/2*k*d^2 + m*g*(h2-h1)=.5*M*v^2

 

[BlueDevil14]

I just keep getting 4.65. I cannot figure out what minor mistake I am making. Its killing me though.

 

[gneill]

Try starting with the gravitational potential energy zero reference set at the spring's compressed state. When the spring is released and reaches its relaxed position, the KE available will be the difference between the stored spring energy and the energy 'sapped' by the various masses rising to the relaxed position.

Call the stored spring energy PS, the change in gravitation potential energy PE, the kinetic energies KS, KP, and KB for the spring, piston, and ball. Then

PS - PE = KS + KP + KB

Can you write expressions for each of the energies?

 

[BlueDevil14]

KS= 1/2 m_s *v^2 where m_s=effective mass of the spring (.017 kg)
KP= 1/2 m_p *v^2 where m_p=mass of piston (.050 kg)
KB= 1/2 m_b *v^2 where m_b=mass of ball (.010 kg)
PS= 1/2 k*d^2 where k=constant (50 N/m), d=displacement (.18 m)
PE= 9.81*d*m_all where m_all= m_s+m_p+m_b

when solving now I get 4.20 m/s

Am I correct in using effective mass for PE?

 

[gneill]

While the ball and piston rise 18cm, the center of mass of the spring will rise less (check the positions if its center of mass before and after).

Use the actual mass of the spring when working out its change in PE, since you'll be working with the change in height of its center of mass.

 

[BlueDevil14]

I changed my PE to reflect center of mass so it is now:

[9.18*.18*(mP+mB)]+[9.8*.09*(.050)]

so v= 4.15 REALLY close to the correct answer

 

[gneill]

Well, to my eye it looks like your calculations are good (I assume that the "9.18*" in your PE expression is a typo for "9.8").

I can't think of anything that's missing, unless the relaxed position of the spring is taken to be with only the mass of the piston (imagine the spring being compressed 18cm from the rest position with just the mass of the piston resting on it, being latched in position, then the ball is placed on top). This would produce a slightly lower 'relaxed' height for the combined mass.

 

[BlueDevil14]

Yes, the 9.18 is a typo for 9.81

Thanks for the help, I think I understand the concepts.