HELP!!! Here's a tough physics problem?

Question

I don't have any idea how to solve this, and this is for my test tomorrow.. please.. give me some ideas or anything you can offer..
Here's the problem:

So there's a block that is hanging by two cords on each end of the block....
the lengths of the cords are described by the letter "l"...
The block's mass is "M"...

so here's the scenerio.. the block hangs.. then, there's this bullet with a mass (m) that hits the block with such speed (v).. and gets implanted into the block.. the impact of the bullet.. made the block go up by a certain height (h)...

the problem is something like that...
find the height after it impacts...
then find the speed of the bullet...
then find the tension of the cord...

there are no numbers.. just letters.. solve this problem any which way you want...

good luck!! please help me!!
if you need more information... that I forgot to imput.. please.. let me know.. thanks!!

Oh I forgot to add something...
when the block gets hit..
it moves a certain height right?.. and it makes an angle theta....

i just forgot to mention that.. sorry..

Answer 1

This is a classic example of a dissipative collision in which conservation of momentum can be used for analysis, but conservation of energy during the collision cannot be invoked because the energy goes into inaccessible forms such as internal energy. After the collision, conservation of energy can be used in the swing of the combined masses upward, since the gravitational potential energy is conservative.


The amount of energy delivered by the bullet can be measured by how high the pendulum's mass is elevated as the pendulum rotates. However not all of the energy from the bullet is transformed to extra potential energy for the pendulum, some is used as heat and deformation energy. But the momentum of the system is conserved and we can use energy conservation after the bullet has hit the pendulum.

Let,
v denote the speed of the bullet
u the speed of the system after the collision

Kinetic energy of the bullet+block AFTER the bullet is embedded=KE
= (1/2) * [ m_bul + M] * u^2

NOTE: u is the velocity AFTER embedding.

Potential energy increase of the bullet+block
=PE
=[m_bul + M] *g *h
g= acceleration due to gravity
=9.81m/sec^2

Sioving for u we get
u = sqrt(2 *g *h)
where, sqrt() means square root of.

Now, although the energy of the bullet when it is shot is partly converted into heat and sound, momentum of th esystem (bullet+block) is always conserved as there is no external force acting in the HORIZONTAL direction. Of course, momentum along the vertical direction is not conserved.

Hence,
p_horiz_before = momentum of the bullet +block before hitting the block along the horizontal direction

p_horiz_after = momentum of the bullet +block after hitting the block along the horizontal direction

Dropping the _horiz subscript, we get
p_before = p_after
or, m_bul * v = [m_bul +M] * u
or, v = ( [m_bul +M]/m_bul ) * sqrt(2 *g *h)

------------------------------
Now in your problem:
l = length of the cords
theta = 5 degrees = angular displacement of the cords

Q1.find the height after it impacts...

After the block has reached the highest point, the end of the cord will be below the hanging point a distance
l*cos(theta)

Draw a picture whic I can not do here.

Hence , the vertical displacement of the block is
h = l - l*cos(theta) = l* [ 1-cos(theta) ]

Hence you can find h.

Q2. then find the speed of the bullet...

v = v = ( [m_bul +M]/m_bul ) * sqrt(2 *g *h)

Q3. then find the tension of the cord

The weight of the block +bullet is

g* [m_bul +M]

which acts directly downward (due to gravity)
Take a component of the weight along the direction of the cord

Tension on each cord = (1/2) * g* [m_bul +M] * cos(theta)

DONE!!

Cheers

Answer 2

Take the momentum of the Bullet P=m*v

Block momentum after the bullet hits it will be the same (momentum conservation) So you have M*V=m*v, more precisely it is (M+m)*V=m*v because the bullet now is part of the block. Here V is the Block with bullet velocity.

That leads you to V=m*v/(M+m) and kinetic energy of this block & bullet is K=(M+m)*sq(V) as they move upwards that Kinetic energy becomes potential energy and eventually when it stops all of the kinetic energy has transformed to potential energy whose formula is U=(M+m)*g*h (g is gravity and h the height).

So simply equate K=U keeping in mind what V is from momentum conservation and then you can solve for height. Dont be affraid of letters they are just a box with some number inside. That makes them more useful than numbers themselves..

Tension in the string can be calculated with:

T-(M+m)*g=(m+M)*Sq(V)/l That is Newtons' second law applied to circular movement

Answer 3

I could solve this...but takes a lot of different formulas to solve.
You have kinetic energy of the bullet. F=MA
A is acceleration.
Then you have angular momentum.
The F lift the block similar to a pendulum.
You have to equate the lifting height of the block with the F of the bullet.

Good luck! Bob

Answer 4

I have some limited physics knowledge, but I'd start by dinging the momentum of the bullet, which is it's speed times its mass. Then, we know that momentum will be transferred to the hanging block, so divide the momentum of the block by the mass of the block to get its speed. Then, know that gravity will slow that block down by 9.8 m/s^2.

That's about where my knowledge ends. Good luck with this, it sounds damn tough

Answer 5

I think it's necessary to know the orientation of the cords, you know, if they're totally vertical or having some angle. Are the two cords united in some point over the block? or are they hanging from diferent points? That info is necesary ...I guess. It seems it's a problem of a kind of "péndulo" (how do you say that in English? ...I'm from mexico) ... you know, something that oscilates hanging from a point. With the info of the angles of the cords, and the basic ecuations of the oscilating movement, you could solve this.

Answer 6

The enrgy of the block is given by E=Mc^2.
When Enstein discovered this law he became famous.

The bullet hits the block with velocity
v = √2kT/3m

The answer is 17.234453455 inch.
Convert to meters.