# Spring projectil wierd problem

1. Dec 8, 2009

### joemama69

1. The problem statement, all variables and given/known data

A block of mass M = 1.0 kg contains a coiled spring and a ball of mass
m = 0.25 kg. The spring is released when the block-ball system is at rest
at the edge of a frictionless plane inclined 200 from the horizontal (see
diagram). The ball, initially 2.0 m above a horizontal floor, strikes the floor
a horizontal distance of 3.0 m from the release point.

a) How far along the plane does the block move before coming to rest?

b) How much energy was initially stored in the spring?

2. Relevant equations

3. The attempt at a solution

Im just confused on what is happening in the problem. Does the block move back up the incline when the ball is pushed out. What does it mean how far does it travel. Its initaly at rest right, or does it slide down the hill and gain some speed then suddenly stop at the end and transfer the momentum to the projectile?

File size:
19.8 KB
Views:
86
2. Dec 8, 2009

### AEM

This is a nice problem! You can use all kinds of good stuff. First, you have conservation of momentum that will tell you something about the velocities of the ball and block immediately after release of the spring. Then you have projectile motion and the location where the ball hits the ground. That can be used to find the initial velocity of the ball. Oh, that allows you to calculate the initial velocity of the block (plus spring?? it doesn't say whether the spring is attached to the block, let's assume the spring is weightless). So then you have conservation of energy since it's a frictionless plane. That will allow you to figure out how high up the block goes. I guess that's enough hints to get you started.

3. Dec 9, 2009

### joemama69

Im still not sure I understanding the time line of events. Is this correct

The whole block ball spring thing is at the top of the incline when it slides down the incline and immediatly stops at the bottom. the momentum is then transfered to the ball and it also recieves some more energy from the spring. and it shoots out like a projectile.

so I would solve for the initial velocity of the projectile v.

and then use conservatio of momentum and energys to find the initial height that the whol system was released from at the very beggingn.

is this correct. wouldnt i have to know more about the spring to know how much energy it contributed to the projectile.

4. Dec 9, 2009

### AEM

No, you want to start with the whole assembly (block, ball, and spring) at the bottom of the incline as shown in your picture. It is at rest so the initial momentum is zero. The spring is compressed, however, and when released it ejects the ball.

Step 1: Write down conservation of momentum equation and set aside.

Step 2: Solve projectile equations to find velocity of the ball leaving the block

Step 3: Use result in conservation of momentum equation to find velocity of block.

Step 4: Use conservation of energy ( KE --> PE) to find height block slides to

Step 5: Use trig to find the distance up the incline the block slides

So, no, you don't need to know details about the spring to know the energy it contained when compressed, but that was a good thought.

5. Dec 9, 2009

### joemama69

ok so the thing is sitting at rest at the end of the inclined plane, and after it shoots there is kind of like a recoil actiong like a gun and it pushes the block up the plane

6. Dec 9, 2009

### AEM

You've got the idea!

7. Dec 9, 2009

### joemama69

i got the velocity to be 17.227 in the direction -20 degrees below the horizontal

so now i do momentum

it i initial at rest = 0

0 = mv1 - Mv2

i need to incorporate the incline. do ihave to find the x compent of the initial velocity and use that as well as find the component of mass and gravity in the same direction when i do the momentum

8. Dec 9, 2009

### AEM

You're on the right track. Now, because you have the velocity of the ball in the direction you gave, why don't you take a coordinate system with the x-axis along the inclined plane and the y-axis perpendicular to it. Then you can just plug your velocity into your (correct) momentum equation and find the velocity of the block.

That allows you to calculate the kinetic energy. That KE --> PE as the block goes up the plane. When it reaches the top of its motion the KE has entirely gone into PE. (We're setting the reference level (PE=0) to be where the block starts out.) So then you can find the height at which the block stopped.

9. Dec 9, 2009

### joemama69

0 = mv1 - Mv2 = .25(17.227) = v2 = 4.307 m/s

so i take that to the energy equation

1/2Mv22 = Mgh, i find the height and relate it to the incline using trig

10. Dec 9, 2009

### AEM

You've got it!

11. Dec 9, 2009

### AEM

Now remember: break all your problems into simple pieces and solve them one piece at a time.

12. Dec 9, 2009

### joemama69

thanks alot, it seemed very confusing but you cleared it up for me. i have two other problems i posted the other day that no one has commened on . could you try to help me out with those

13. Dec 9, 2009

### Mr confusion

friends,sorry to intervene,but i am having a problem with b.) the initial energy stored in spring
now by energy conservation, i can say that it will be equal to 1/2 Mv block^2+ 1/2 mv ball^2
what velocities will i use ? the ones immediately after impact or some time later?

14. Dec 9, 2009

### AEM

Use the velocities right after the ball has been released.

15. Dec 10, 2009

### Mr confusion

first of all, many thanks for the reply. i have just one more doubt regarding this.
we take x axis along incline, right?
then we conserve momentum along the incline, right?
but my book says to use momentum conservation only when no external foce acts on the system. but here a external force IS acting on the system and that is(m ball + M block)g sin theta. please tell me how can we conserve momentum in that direction then?
i may be wrong, though. any help will be heartily welcomed,

16. Dec 10, 2009

### AEM

That's a good question. The answer is that gravity is operation on both masses simultaneously AND gravitational mass is the same as inertial mass. That is, if we write Newton's law of motion F = Ma where F is the gravitational force due to the earth,

$$Ma = G \frac{ M_e M}{r^2}$$,

we don't hesitate to cancel the M on the left hand side and the M on the right hand side. Even though the M on the left arises as the resistance of the object to changing its state of motion, and the M on the right hand side arises as the object's response to gravity. The equivalence of gravitational mass and inertial mass has been demonstrated to 1 part in 10^(-12), a very strong result.

That is why we can ignore gravity as an external force in using conservation of momentum for problems like these. What your book was referring to was other types of external forces (an attached spring, a magnetic force, etc) in which case you would have to be careful to examine exactly what's happening.

17. Dec 10, 2009

### Mr confusion

understood. thanks.