Inelastic collision and spring compression problems

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*A 50 g ball of clay traveling at speed v_i hits and sticks to a 1.0 kg block sitting at rest on a frictionless surface.

1.What is the speed of the block after the collision?

2. What percentage of the ball's initial energy is "lost"?

*A vertical spring with k = 490 N/m is standing on the ground. You are holding a 5.0 kg block just above the spring, not quite touching it.

1.How far does the spring compress if you let go of the block suddenly?

2.How far does the spring compress if you slowly lower the block to the point where you can remove your hand without disturbing it?

i would appreciate if u would help me with these problems or try to guide me at least ^^
 

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For the first one, law of conservation of mometum.

[tex]m_1v_1+m_2v_2=m_1v'_1_m_2v'_2[/tex]
 
thnx^^ anybody know how to do the second one?
 
first part
mv=m2v2
v2=mv/m2
v2=.05*v/(.05+1)

2)calculate the kinetic energy after hitting the brick using the calculated velocity
1/2mv^2
1/2*(1+.05)*(the above calculated velocity)^2

then reduce it from the starting kinetic energy, then u get the energy that was wasted then u will be able to fine the percentage

the second part of the question is
F=Ke
50=490*e
e=50/490

the other part i still havnt figurd out
 
xyoonboixkpx said:
thnx^^ anybody know how to do the second one?

You know the initial speed and in part one found the final speed. What are the initial and final kinetic energies?
 
For the second part, the spring is in equilibrium. The forces acting are the spring force, and the weight of the object.
 
For the second part, I think it would just be,

Fs = Fg, where Fs is the force of the spring and Fg is the force due to gravity.

So,

kx = mg,

where k is the spring constant, m is the mass of the brick in kg, g is the acceleration due to gravity, and x is the displacement.

Solve for x. The numeric sign of the displacement is relative. You could make it positive or negative, but typically the force of a spring is written as,

Fs = -kx.