# Homework Help: Stuck on some problems

1. Mar 11, 2005

### ness9660

1. A 66.3 g mass is attached to the end of
an unstressed vertical spring (of constant
63.5 Nm) and then dropped.
The acceleration of gravity is 9.8 m/s2 :
What is its maximum speed? Answer in
units of m/s.

Im not quite sure that I understand what this problem is saying. Is the block hanging from the spring? Is it on top of the spring? Im not sure how this system is setup, but beyond that it is a harmonic motion problem, correct?

2.A(n) 1.6 kg object moving at a speed of
6.5 m/s strikes a(n) 1.2 kg object initially
at rest. Immediately after the collision, the
1.6 kg object has a velocity of 0.88 m/s di-
rected 46 degrees from its initial line of motion.
What is the speed of the 1.2 kg object
immediately after the collision? Answer in
units of m/s.

Ive been doing it like this:
X direction: m1i*v1i=m1i*v1f*cos(46) + m2*v2f*cos(46)

Y direction: m1*v1i=m1*v1f*sin(46) + m2*v2f*sin(46)

and then v2f= sqrt(x^2 + y^2)

However this is apparently wrong, am I close in my approach to this problem?

2. Mar 11, 2005

### Gamma

I believe the other end of the spring is attached to a ceiling. If that is the case, then you can apply newton's second law to a point where the mass has moved a distance x from it's equilibrium position.

mg - kx = m d^2 x/ dt^2 and solve for dx/dt. That will lead you to the maximum velocity.

3. Mar 11, 2005

### Gamma

You have assumed that mass m2 travells in the same direction as m1 after the collision. This assumption is wrong. Assume that m2 travell in a direction theta with respect to the original direction of m1.

m1i*v1i=m1i*v1f*cos(46) + m2*v2f*cos(theta)
0 =m1*v1f*sin(46) - m2*v2f*sin(theta)

solve for theta.