How Do You Calculate Momentum in Different Collision Scenarios?

In summary, Three questions were discussed regarding momentum and the conservation of energy. The first question involved the collision of a putty and a ball, with the conclusion that both objects move with a momentum of 3Ns. The second question involved a bullet striking a metal plate and ricocheting off at an angle, with the solution involving the use of vector addition and geometry. The third question involved a bullet embedding itself in a wood pendulum and causing it to rise, with the solution involving the conservation of energy and calculating the initial velocity of the bullet.
  • #1
NoHeart
28
0
I have three questions that I'm having a hard time with. Any help would be greatly appreciated...

easiest one first:

1. A 3kg ball of putty moving at 1m/s collides and sticks to a 2kg ball at rest. The putty and the ball then move with a momentum of what?

so far i have initial momentum of putty at 3Ns (3kg*1m/s) and then I'm not quite sure... initial momentum of other ball is zero? (2kg*0m/s=0Ns)
so do they both keep moving with a momentum of 3Ns? or is the force not enough for them to move, and it's 0Ns? I'm thinking the momentum cannot remain constant because the mass has changed...so would it be correct to say 3kg-2kg=1kg, and that momentum would equal 1Ns? I'm thinking this is the answer (1Ns) but I'm not sure if I'm grasping the concept correctly...

2. a 20g bullet travels at 350m/s, strikes metal plate at an angle of 30, bullet ricochets off at same angle with a speed of 320m/s- What is the magnitude of the Impulse the wall gives to the bullet?

this is giving me a lot of trouble, I'm not sure what to do with the angles...the diagram shown has the bullet traveling at an angle pointing up towards the wall, and ricocheting off upwards, so the angles are in reference to the y-axis...
if impulse is a change in momentum, i have initial momentum at 7Ns (20g*350m/s) and final momentum at 6.4Ns, so the impulse would be
6.4-7= -0.6Ns
that's not anything close to what i think the answer is, i know i should be using sin or cos30 at some point, but where? and is the initial momentum going to be negative because it is coming from the negative x-axis?
i am so confused...

3. a 10.5g bullet strikes a 3kg wood pendulum and imbeds in the wood. the pendulum rises by 0.22m after impact, What was the initial velocity of the bullet?

okay, so lost. the initial momentum has to equal the final momentum, right? but how do i find momentum without velocity? i could find the final momentum using the distance of 0.22m, but i don't know the time. where do i even start here?
 
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  • #2
NoHeart said:
I have three questions that I'm having a hard time with. Any help would be greatly appreciated...

easiest one first:

1. A 3kg ball of putty moving at 1m/s collides and sticks to a 2kg ball at rest. The putty and the ball then move with a momentum of what?

so far i have initial momentum of putty at 3Ns (3kg*1m/s) and then I'm not quite sure... initial momentum of other ball is zero? (2kg*0m/s=0Ns)
so do they both keep moving with a momentum of 3Ns?

Yes. Momentum is conserved. They keep moving with 3Ns.

2. a 20g bullet travels at 350m/s, strikes metal plate at an angle of 30, bullet ricochets off at same angle with a speed of 320m/s- What is the magnitude of the Impulse the wall gives to the bullet?

Suppose the initial momentum is vector a. The final momentum is vector b. We want the magnitude of the vector b-a (use the parallelogram rule for adding vectors...). Draw a sketch of the vector b-a (sketch b... add -a). You'll need to use a little geometry to get the magnitude of b-a.

3. a 10.5g bullet strikes a 3kg wood pendulum and imbeds in the wood. the pendulum rises by 0.22m after impact, What was the initial velocity of the bullet?

okay, so lost. the initial momentum has to equal the final momentum, right? but how do i find momentum without velocity? i could find the final momentum using the distance of 0.22m, but i don't know the time. where do i even start here?

There are 2 stages to this problem. First there is the bullet striking and embedding itself in the wood... here momentum is conserved (momentum is in the horizontal direction and there are no forces acting in the horizontal direction).

In the second stage(pendulum moving upwards with the bullet) momentum is not conserved because we have an external force (gravity) acting against the motion of the pendulum. But we have energy conservation here.

Try working backwards from stage 2 to stage 1.
 
  • #3
so for the 3rd one, i need to deal with energy...kinetic and potential energy are conserved...? so i can say kinetic energy after the collision is1/2mv^2 and that is equal to the potential energy of the pendulum after impact, which is mgh...so m=10.5g+3kg=3.0105 and h=0.22m
this leads to final velocity at 2.08m/s
final momentum is 0.0105kg*2.08m/s+3kg*2.08m/s=6.26m/s
how do i use final momentum to get initial velocity if momentum is not conserved? are you sure the initial momentum isn't equal to the final momentum?
 
  • #4
NoHeart said:
so for the 3rd one, i need to deal with energy...kinetic and potential energy are conserved...? so i can say kinetic energy after the collision is1/2mv^2 and that is equal to the potential energy of the pendulum after impact, which is mgh...so m=10.5g+3kg=3.0105 and h=0.22m
this leads to final velocity at 2.08m/s
final momentum is 0.0105kg*2.08m/s+3kg*2.08m/s=6.26m/s
how do i use final momentum to get initial velocity if momentum is not conserved? are you sure the initial momentum isn't equal to the final momentum?

Good work!

Yes, you are right momentum is conserved (this is stage 1 now - the collision). You can calculate initial velocity just like you said.

What I meant was that you can't use conservation of momentum in stage 2 (when the pendulum goes up).
 
  • #5
thank you and goodnight

thank you so much! i ended up with the answer in terms of acceleration though, 596.2m^2/s
i think that'll work, the exact question was "how fast was the bullet moving" velocity wasn't specifically mentioned.
now I'm going to go study vectors...
 
  • #6
NoHeart said:
thank you so much! i ended up with the answer in terms of acceleration though, 596.2m^2/s
i think that'll work, the exact question was "how fast was the bullet moving" velocity wasn't specifically mentioned.
now I'm going to go study vectors...

Careful with your units. I noticed that you wrote 6.26 m/s instead of 6.26kg*m/s for your momentum.

When you set 0.0105 kg * v = 6.26 kg*m/s

v comes out to 596.2 m/s (units come out to m/s)

Also, another thing... try not to round much until you need your final answer... I got 595.4 m/s when I didn't round till the end.

Good luck.
 
  • #7
thanks again,LP, that certainly makes a difference in my answer...

for #2, in my AC electronics class a couple semesters ago we learned how to convert polar coordinates to rectangular- can i apply this to the problem? if I'm starting with the two polar coordinates of 350m/s at an angle of 30(or is it 150?) and 320m/s at an angle of 30, should i convert them to rectangular and add them? and is that answer the magnitude?
 
  • #8
NoHeart said:
thanks again,LP, that certainly makes a difference in my answer...

for #2, in my AC electronics class a couple semesters ago we learned how to convert polar coordinates to rectangular- can i apply this to the problem? if I'm starting with the two polar coordinates of 350m/s at an angle of 30(or is it 150?) and 320m/s at an angle of 30, should i convert them to rectangular and add them? and is that answer the magnitude?

Hi NoHeart. Yes, you can certainly do it this way. I'd sketch in the vertical and horizontal components of each vector. This way you can see the angles, and you can get the vertical and horizontal component by using the sine or cosine. You just have two right triangles.

Essentially you are getting the horizontal and vertical components of the final momentum... and also the horizontal and vertical components of the initial momentum.

Then the horizontal component of the impulse is the horizontal component of the final momentum minus the horizontal component of the initial momentum. And the vertical component of the impulse is the vertical component of the final momentum minus the vertical component of the initial momentum.

If the initial velocity is upwards to the right (wall is on the right side) making an angle of 30 degrees with the vertical... then the horizontal component of momentum is mvsin(30) and the vertical component is mvcos(30). Do you see this with the right triangle? Be careful regarding signs.

When you get the horizontal and vertical components of the impulse use the pythagorean theorem to get the magnitude.

Hope this helps.
 
  • #9
For question number 1, how would you find the velocity of the two bodies after collision?
 
  • #10
the_d said:
For question number 1, how would you find the velocity of the two bodies after collision?

We have the momentum = 3Ns or 3kg*m/s

The total mass of the ball and putty=5 kg.

Momentum = m v

So v=momentum/m = 3/5 = 0.6 m/s
 
  • #11
still have a problem

my way i get .59 at an angle of -30
your way i get .59
this is a multiple choice question and my options are 300 Ns, .30Ns, .52Ns, and 6.7Ns

the way i got .59 was
a=mvsin30=3.5
b=mvsin30=3.2

a=mvcos30=6.06
b=mvcos30=5.54

Impulse a= 3.2-3.5=-.3
Impulse b= 5.54-6.06= -.52

Magnitude^2= -.3^2+-.52^2
Magnitude=.59

what am i doing wrong?
 
  • #12
NoHeart said:
my way i get .59 at an angle of -30
your way i get .59
this is a multiple choice question and my options are 300 Ns, .30Ns, .52Ns, and 6.7Ns

the way i got .59 was
a=mvsin30=3.5
b=mvsin30=3.2

a=mvcos30=6.06
b=mvcos30=5.54

Impulse a= 3.2-3.5=-.3
Impulse b= 5.54-6.06= -.52

Magnitude^2= -.3^2+-.52^2
Magnitude=.59

what am i doing wrong?

Hi Noheart. The initial horizontal momentum is 3.5 Ns. The final horizontal momentum is -3.2Ns (since it's in the opposite direction). So the horizontal component is -3.2-3.5=-6.7

The vertical impulse is -0.52 as you calculated.

magnitude^2 = (-6.7)^2+(-0.52)^2
magnitude = 6.72

So 6.7 is your choice.
 
  • #13
will you marry me?
just kidding...now i see what you meant about being careful of signs...i am so happy that i actually understand this now!
 
  • #14
NoHeart said:
will you marry me?
just kidding...now i see what you meant about being careful of signs...i am so happy that i actually understand this now!

Why aren't the marriage proposals ever serious?! :smile:

:biggrin: Glad to help!
 

Related to How Do You Calculate Momentum in Different Collision Scenarios?

1. What is the difference between impulse and momentum?

Impulse refers to the product of force and time, while momentum refers to the product of mass and velocity. In other words, impulse is a measure of how much force is applied over a certain period of time, while momentum is a measure of an object's motion.

2. How are impulse and momentum related?

Impulse and momentum are related through Newton's Second Law of Motion, which states that the change in an object's momentum is equal to the impulse applied to it. This means that the larger the impulse, the larger the change in momentum, and vice versa.

3. Can impulse and momentum be negative?

Yes, both impulse and momentum can be negative. Negative impulse occurs when a force is applied in the opposite direction of an object's motion, while negative momentum occurs when an object is moving in the opposite direction of its initial velocity.

4. How does impulse affect an object's motion?

Impulse can change an object's motion by either increasing or decreasing its velocity. When a large impulse is applied to an object in a short amount of time, it can cause a large change in velocity, resulting in a change in the object's motion.

5. What are some real-world applications of impulse and momentum?

Impulse and momentum are applicable in many real-world scenarios, such as car collisions, sports, and rocket propulsion. In a car collision, the impulse of the impact can determine the amount of damage done to the vehicles and passengers. In sports, understanding momentum can help athletes improve their performance, such as in long jumping or throwing events. In rocket propulsion, the impulse of the fuel being ejected determines the momentum of the rocket and its ability to overcome the force of gravity and achieve lift-off.

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