Having a hard time grasping the concept with momentum

[TeaWay]

So I'm struggling trying to grasp the concept of momentum..due to my physics teacher going WAAAAAAAY too fast && i asked for help he didn't help much at all :\..so i asked my previous physics teacher i had before transferring classes because the school " said so " but it was a bit still confusing ; - ; ...

can someone help me break down the concept a bit and help guide me through this question i am about to provide... before my big exams ): .

A 4kg ball moving at 2 m/s to the right collides head on with a 2kg ball moving at 1 m/s to the left .

a) If the collision is perfectly inelastic, what are the speed and direction of the balls after the collision.

My Work:

p = (m₁ + m₂ ) v₃
v₃= -m₁v₂+pƒ
m₂

p=4kg(2m/s) p=2kg(-1m/s)
p=8kg*m/s p= -2kg*m/s

p initial = 8kg*m/s - 2kg*m/s
p initial = 6kg*m/s

pƒ =(4kg+2kg)v₃
p initial = 6kg*m/s
v₃= 6kg*m/s
6kg
v₃ = 1 m/s

p initial = pƒ (final)

b) if the 4kg ball comes to rest after the collision, what are the speed and direction of the 2kg ball after the collision?

p=4kg(2m/s)
p=8kg*m/s

p initial = p₁ + p₂
p initial = 8kg*m/s - 2kg*m/s
p initial = 6kg*m/s = m₂v₄ = pƒ(final)

c) if the 4kg ball continues to move to the right at 1 m/s after the collision, what are the speed and direction of the 2kg ball after the collision?

d) how much momentum was transferred during this collision?

e) what is the impulse of the collision ?

f) what was the total kinetic energry of the two balls before the collision ?

g) what was the total kinetic energry of the two balls after the collision

h)was the collision elastic or inelastic ? why ?

sorry if i did very little work :\ I am trying to stuff this all in my brain asap because i just finished ch 5&6 a week ago and starting ch 7 and now into 8 soon ... my teacher is going too fast and he tells the class he will not slow down so basically he says " if your behind . we will leave you behind.." I am quite worried :\ i need to pass this class so i can do 4 years of science to set my foundation then when i graduate ..ASTRONOMY HERE I COME O:< ! but i feel discourage :\.

 

[Schr0d1ng3r]

Your answer for a is correct.

b) p(system) = m1v1 + m2v2 = 6kgm/s

(2kg)*v2(final) = 6kgm/s

v2(final) = 3m/s

c) p1i + p2i = p1f + p2f

8kgm/s - 2kgm/s = 4kgm/s + 2kg*v2f

v2f = 1m/s

d) (Delta)p = pf - pi

(Delta)p = 2kgm/s + 2kgm/s

(Delta)p = 4kgm/s

e) i = (Delta)p/(Delta)t

^You are going to need to know how long the collision took for this part

f) T = 1/2 * m1(v1^2) + 1/2 * m2(v2^2)

I'm sick of plugging in numbers at this point, but its straightforward enough

g) There are no external forces, so the kinetic energy should remain constant

h) Since the balls have the same final velocity, they must have stuck together, meaning that the collision was inelastic​

 

[TeaWay]

Your answer for a is correct.

b) p(system) = m1v1 + m2v2 = 6kgm/s

(2kg)*v2(final) = 6kgm/s

v2(final) = 3m/s

c) p1i + p2i = p1f + p2f

8kgm/s - 2kgm/s = 4kgm/s + 2kg*v2f

v2f = 1m/s

d) (Delta)p = pf - pi

(Delta)p = 2kgm/s + 2kgm/s

(Delta)p = 4kgm/s

e) i = (Delta)p/(Delta)t

^You are going to need to know how long the collision took for this part

f) T = 1/2 * m1(v1^2) + 1/2 * m2(v2^2)

I'm sick of plugging in numbers at this point, but its straightforward enough

g) There are no external forces, so the kinetic energy should remain constant

h) Since the balls have the same final velocity, they must have stuck together, meaning that the collision was inelastic​

thank you very much ; - ; ... now i shall study with the answers you have provided :D !​

 

[Schr0d1ng3r]

No problem. Good luck!

 

[TeaWay]

No problem. Good luck!

oh may i ask a question ? .

c) p1i + p2i = p1f + p2f

8kgm/s - 2kgm/s = 4kgm/s + 2kg*v2f

how did you get the 4kg*m/s && 2kg*v2f ? i just want to double check aha.. ; - ; so i don't get the whole concept wrong and whatnot.

and for this

(Delta)p = pf - pi

(Delta)p = 2kgm/s + 2kgm/s

(Delta)p = 4kgm/s

p final and p initial will be the same right ? O:

i think.. ahh i forgot what i learned in class today about this :\

 

[Schr0d1ng3r]

for c)

4kgm/s comes from the final momentum of the 4kg ball (4kg*1m/s) and 2kg*v2f comes from the final momentum of the 2kg ball, where v2f is its final velocity

for d)

They end up being negatives of each other in this question, but won't necessarily every time. pf is the final momentum of the object in question, and pi is its initial momentum

 

[TeaWay]

for c)

4kgm/s comes from the final momentum of the 4kg ball (4kg*1m/s) and 2kg*v2f comes from the final momentum of the 2kg ball, where v2f is its final velocity

for d)

They end up being negatives of each other in this question, but won't necessarily every time. pf is the final momentum of the object in question, and pi is its initial momentum

thank you very much ^^ ! you helped me so much (: !

 

[Schr0d1ng3r]

No worries. I hope that you ace the momentum part of your exam at very least ;) And if you need help with anything else, just ask.