# (Momentum I Think)

Could someone help me with this problem? I'd greatly appreciate it!

A block of mass M = 1.00 kg contains a coiled spring and a ball of mass m = 0.250 kg.
The spring is released when the block-ball system is at rest on the edge of a frictionless
plane inclined 20 degrees from the horizontal. The ball, initially 2.00 m above the horizontal
floor, strikes the floor a horizontal distance of 3.00 m from the release point. How far
along the plane does the block move? How much energy was initially stored within the
spring?

(Picture attached)

I've tried summing forces, conservation of enrgy, conservation of momentum, and I can't work it out. Any little hint helps! Thanks!

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Doc Al
Mentor
Show what you've done so far. You'll need projectile motion, conservation of energy, and conservation of momentum to crack this one.

For conservation of energy I was looking at the block and had:
(1/2)kx^2=mgh
K is the spring constant, x the spring compression, m mass, g gravity, and h the hieght the block will get on the slope by the time its kenetic energy is 0. Still I have three unknowns.

For the ball:
(1/2)kx^2+2mg=(1/2)mv^2 (spring potential + gravitational potential=kenetic)

In these two equations k and x are the same, but I still have v and h.

I tried using this momentum equation, but it could be wrong:
mv(block)+mv(Ball)=mu(Block)+mu(Ball)
v equals initial velocity, and u final. The initials were 0, so I had:
mu(block)=-mu(Ball) but I'm not sure if this relationship is worth anything, or even correct.

Doc Al
Mentor
For conservation of energy I was looking at the block and had:
(1/2)kx^2=mgh
K is the spring constant, x the spring compression, m mass, g gravity, and h the hieght the block will get on the slope by the time its kenetic energy is 0. Still I have three unknowns.
A mistake here is thinking that all the spring energy goes into the block. What about the ball?

For the ball:
(1/2)kx^2+2mg=(1/2)mv^2 (spring potential + gravitational potential=kenetic)

In these two equations k and x are the same, but I still have v and h.
Think of the expanding spring as an "explosion": You start out with spring PE and end up with KE of block and ball. (After the spring expansion the ball shoots off and the block rises. Treat that separately.)

I tried using this momentum equation, but it could be wrong:
mv(block)+mv(Ball)=mu(Block)+mu(Ball)
v equals initial velocity, and u final. The initials were 0, so I had:
mu(block)=-mu(Ball) but I'm not sure if this relationship is worth anything, or even correct.
You'll definitely need this. Use it to find the relationship between the initial speeds (immediately after the spring expansion) of block and ball.

Hint: Use what you know about projectile motion to figure out the initial speed of the ball after the spring expands.

The thing I can't get is the projectile motion, and it should be the easiest part! I feel so stupid.
I tried using x=x(o)+v(o)t+(1/2)at^2
In the Y direction I had:
2=-vsin(20)+(.5*-9.8)t^2 but with this i dont know v or t and I dont know how to set up an x equation as it has an acceleration applied.

Wait I was being really stupid. The x acceleration is 0 isn't it? now the problem is I don't know what to do with that velocity!

Doc Al
Mentor
The thing I can't get is the projectile motion, and it should be the easiest part! I feel so stupid.
I tried using x=x(o)+v(o)t+(1/2)at^2
As you realized, the horizontal acceleration is 0. Fix this equation, putting in the correct initial speed (horizontal component, in terms of v and the angle) and final position.
In the Y direction I had:
2=-vsin(20)+(.5*-9.8)t^2 but with this i dont know v or t and I dont know how to set up an x equation as it has an acceleration applied.
This is the right idea. Two errors: That final y position should be -2; you left out a factor a t.

You'll have two equations and two unknowns (v and t): Solve!

Wait I was being really stupid. The x acceleration is 0 isn't it? now the problem is I don't know what to do with that velocity!
See above.

Thanks for all your help I think I solved it.