Question

A spring of equilibrium (un-stretched) length L0 is hung vertically from one end. A mass M is attached to the other end of the spring and lowered so that the mass hangs stationary with the spring stretched a distance ΔL.

The position of the bottom end of the un-stretched spring is defined as y=0 and shown by the (upper) blue line in the figure. The position of the end of the stretched spring is shown by the (lower) red line in the picture.

A projectile with mass m is fired vertically into the mass on the spring. The projectile has a speed vi when it hits M. The projectile becomes embedded in the larger mass so that they move upwards, compressing the spring, and reaching a maximum height yf. By measuring the maximum height of the mass (and projectile) on the end of the spring, it's possible to determine the initial speed of the projectile.

IMPORTANT: The maximum height, yf, is measured from y=0, the position of the end of the un-stretched spring. If yf is positive, that is a position above y=0, as usual.

In this problem, the "system" is the projectile, the mass, the spring, and the Earth.

Part A: Derive an equation for the spring constant of the spring, K, using (some of) the variables defined above. Think carefully about what you know. You shouldn't need to write much; this should just take two or three lines.

Part B: Derive an equation for the speed of the mass M and projectile m immediately after the projectile is embedded in the mass using (some of) the variables defined above. Again, this should just take two or three lines.

Let M=3.56 kg, m=0.26 kg, L0=0.95 m, ΔL=0.35 m, and yf= 0.70 m.

Part C: Solve for the spring constant, K.

Part D: Write the energy conservation equation for the total change in energy from right after the collision (initial time) until the maximum height of the mass and projectile (final time). This should include all energies for both the initial and final times. Include zeros if an energy (initial or final) is zero.

Part E: Solve the for the value of the initial velocity of the projectile using the values given above

Answer 1

Given, Mass M = 3.56 kg

mass of projectile ( m ) = 0.26 kg

unstreched lenght of spring ( Lo ) = 0.95 m

change in length when mass M is attached ( ) = 0.35 m

maximum height after the projetile embedded ( ) = 0.70 m

acceleration due to gravity ( g ) =

Let the spring constant is k.

It is said to us that when the spring is unstretched i.e. y=0. So, we consider it as the reference point for gravitational potential energy.

(a) When the mass is attached to spring, it will be extended. At , forces on mass M will be balanced.

So, the free body diagram of the mass M is

balancing force in y-direction,

   ......( 1 )

(b) As, the system is the projectile, the mass, the spring, and the Earth. So, there will be no external force.

Hence, the momentum will be conserved.

Let the speed of the mass M and projectile m mmediately after the projectile is embedded in the mass is 'v'.

Applying conservation of momentum in y-direction,

  

   ....( 2 )   

(c) putting the values in equation ( 1), we get

  

  

(d) The total change in energy will be zero because there is no non-conservative force.

Three types of energy are present here i.e. kinetic energy, spring potential energy and gravitational potential energy.

Therefore, equation will be

   ...( 3 )

(e) solving equation ( 3 ) further more and putting the given values,

  

     

  

putting this in equation ( 2 ),

Hence, the initial velocity of the projectile ( ) = 80.66 m/s