Hooke's Law Problem: Finding Maximum Displacement with Inelastic Collision

[sr57]

A 0.2 kg block of wood is attached to a spring with a spring constant k = 25 N/m. The block is initially at rest and the spring is at its equilibrium length aligned along the x-axis. A dart of mass 0.05 kg is thrown at a block of wood, undergoes an inelastic collision and sticks into the block. the initial speed of the dart is 0.1 m/s and is parallel to the x-axis. The maximum displacement of the block from its equilibrium position in m is: ?

Homework Equations

The Attempt at a Solution

I used m1Vi + m2Vi = (m1+ m2) Vf
to find Vf = 0.1 m/s

I know F= k x
I don't know know how you would find F.
Please help

 

[Tedjn]

Please recheck your v_f, because I find something different using your equation. Then, you can try using Hooke's law with conservation of energy (the spring force is a conservative force). Where is the energy from the moving block and dart going when they slow down to a stop?

 

[oedipa maas]

I don't think you have the right value for v_f -

(0.2 kg)*(0 m/s) + (0.05 kg)*(0.1 m/s) = (0.25 kg)*v_f
v_f = 0.02 m/s

Although the collision between the dart and the block is inelastic, the compression of a spring does conserve the sum of the kinetic energy and the elastic potential energy!

 

[sr57]

Thank you I got it. I made a mistake with the calculations b4..but this is the way right?

Since F= PV => F = (mv)v ==> F= 0.25

F= kx
0.25/25 N/m = x
x= 0.01 m

 

[Tedjn]

Why do you say F = pv?

 

[oedipa maas]

F is not equal to the product of momentum and velocity... (you can see this by comparing the units of F (kg*m/s^2) and the units of pv (kg*m/s*m/s)).

It's important to see here that the sum of elastic potential energy and kinetic energy is conserved. So what is the elastic potential energy of a spring compressed distance x?

 

[sr57]

Sorry I meant P = Fv therefore F = P/V and since P=mv, F = mv/v = m ?

 

[oedipa maas]

Whoa, you've just said F=mv/v = m... But force isn't equal to mass!

 

[sr57]

F is not equal to the product of momentum and velocity... (you can see this by comparing the units of F (kg*m/s^2) and the units of pv (kg*m/s*m/s)).

It's important to see here that the sum of elastic potential energy and kinetic energy is conserved. So what is the elastic potential energy of a spring compressed distance x?

So 1/2 mv^2 = 1/2 kx^2
x = 0.01 m

 

[oedipa maas]

Can you show your substitution? There seems to be a mistake there.

 

[sr57]

1/2 (0.25 kg)(0.02 m/s) = 1/2 (25 N/m)(x^2)
x^2 = 0.002 m

I put the answer for the rong question before..sorry abt that :$