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problem 2:
An ice hockey puck slides along the ice at 12m/s. A hockey stick delivers an impulse of 4.0 kg*m/s causing the puck to move off in the opposite direction with the same speed. What is the mass of the puck in kg?
Problem 3:
A small, 100g cart is moving at 1.90m/s on an air track when it collides with a larger, 5.00kg cart at rest. After the collision, the small cart recoils at 0.830m/s.
What is the speed of the large cart after the collision? Answer in m\s.
thank you so much in advance to anyone who helps!

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Do you need to account for friction or do you assume a frictionless surface?
That's why once you're in the game, you're in it for life.
DJAuto

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with no coefficient i'm assuming it's frictionless

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back to page one, still looking for help

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and again

Banned
you need to do with using momentum.
They dont say whether right or left is positive. But looks like it implies that right is positive.
a. So initiall 1.5kg mass moving at 1.1m/s
initial momentum = 1.65kgm/s
2N force acts for 0.5S. Impulse (change in momentum) is +2*0.5 = 1kgm/s
final momentum = 1.65+1 = 2.65kgm/s
Velocity = momentum/mass = 2.65/1.5 = +1.77m/s
b. Velocity is still positive so its still moving to the right.
c. Almost same problem, force and therefore the change in momentum is now negative.
impulse = 2*0.5 = 1kgm/s
final momentum = 1.651 = +0.65kgm/s
Velocity = 0.65/1.5 = 0.433m/s
d. Velocity is still positive so its still moving to the right.

Banned
Originally Posted by zjp6050
problem 2:
An ice hockey puck slides along the ice at 12m/s. A hockey stick delivers an impulse of 4.0 kg*m/s causing the puck to move off in the opposite direction with the same speed. What is the mass of the puck in kg?
Problem 2
let mass = m, initial velocity = 12m/s, final velocity = 12m/s
initial momentum = 12m
final momentum = 12m
change in momentum = impulse = 4kgm/s
therefore:
initial momentum + change in momentum = final momentum
12m  4 = 12m
24m = 4
m = 4/24 = 0.1667Kg= 167g
Last edited by damlurker; 11112008 at 10:35 PM.

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Originally Posted by damlurker
Problem 2
let mass = m, initial velocity = 12m/s, final velocity = 12m/s
initial momentum = 12m
final momentum = 12m
change in momentum = impulse = 4kgm/s
therefore:
initial momentum + change in momentum = final momentum
12m  4 = 12m
24m = 4
m = 4/24 = 0.1667Kh= 167g
if the answer is in kg would that 167g = 1.7kg in 2 sig figs? if so, it was wrong. again i am most likely putting it in wrong.
thank you a ton by the way!

Banned
Originally Posted by zjp6050
Problem 3:
A small, 100g cart is moving at 1.90m/s on an air track when it collides with a larger, 5.00kg cart at rest. After the collision, the small cart recoils at 0.830m/s.
What is the speed of the large cart after the collision? Answer in m\s.
Use law of conservation of momentum
let 100g cart be cart A with Mass Ma=0.1kg with velocity Vinitial=1.9m/s and Vfinal = 0.83m/s
let 5kg cart be cart B with Mass Mb=5kg, with Vinitial=0m/s and Vfinal = V m/s
initial momentum of cart A = Ma*Vinitial = 0.1*1.9=0.19kgm/s
final momentum of cart A = Ma*Vfinal = 0.1*0.83 = 0.083kgm/s
initial momentum of cart B = Mb*0 = 0kgm/s
final momentum of cart B = Mb*V = 5V kgm/s
initial momentum of cart A + initial momentum of cart B = final momentum of cart A + final momentum of cart B
0.19+0 = 0.083 + 5V
V = (0.190.083)/5
V = 0.107/5
V = 0.0214 m/s

Banned
Originally Posted by zjp6050
if the answer is in kg would that 167g = 1.7kg in 2 sig figs? if so, it was wrong. again i am most likely putting it in wrong.
thank you a ton by the way!
that would be 0.17kg to two significant figures.
I didnt double check the numbers but the theory should be right.
Looks like you're course is trying to teach momentum, impulse and the law of conservation of momentum...
I miss this stuff..

Banned
do u still want q1 done
cbf reading replies

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ugh i think i'm putting them in wrong. all answers must be in two sig figs...as i put them in i put...for problem 2: 1.7m\s
problem 3: .021
am i doing something wrong?

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Originally Posted by PeteCraig
do u still want q1 done
cbf reading replies
questions 1 and 2 i have gotten correct. question 3 i just think i'm putting in wrong. do you guys mind if i post a few more questions?

Banned
well **** U im doing it anyway
Impulse = F*dt = m*dv Which is equal to the area under the graph
= 1
So 1 = m * dv
since m = 1.5
1/1.5 = dv
so Final speed = 1/1.5 + 1.1
= 1.76 m s^1
**** U PHYSICS

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lol that was right. you care if i post a couple more?

Banned
Originally Posted by PeteCraig
well **** U im doing it anyway
Impulse = F*dt = m*dv Which is equal to the area under the graph
= 1
So 1 = m * dv
since m = 1.5
1/1.5 = dv
so Final speed = 1/1.5 + 1.1
= 1.76 m s^1
**** U PHYSICS
**** son.. u using calc nao..
physics is serious business..
Last edited by damlurker; 11112008 at 10:49 PM.

Banned
Originally Posted by zjp6050
ugh i think i'm putting them in wrong. all answers must be in two sig figs...as i put them in i put...for problem 2: 1.7m\s
problem 3: .021
am i doing something wrong?
If my calculations are correct, the answers to 2 significant figures:
Problem 2: 0.17Kg
Problem 3 0.021m/s

Banned
Originally Posted by damlurker
**** son.. u using trig nao..
physics is serious business..
trig? I havnt done physics for like 6 yrs and that was all from memory
WAS I CORECT?

Registered User
Originally Posted by damlurker
If my calculations are correct, the answers to 2 significant figures:
Problem 2: 0.17Kg
Problem 3 0.021m/s
problem 3 was incorrect
yes pete, you were

Banned
Originally Posted by zjp6050
problem 3 was incorrect
yes pete, you were
**** U PHYSICS

Coming at you bro
all of them......
bout tree fiddy?
PackersCelticsRedSoxLeafs

Banned
Originally Posted by zjp6050
problem 2:
An ice hockey puck slides along the ice at 12m/s. A hockey stick delivers an impulse of 4.0 kg*m/s causing the puck to move off in the opposite direction with the same speed. What is the mass of the puck in kg?
Problem 3:
A small, 100g cart is moving at 1.90m/s on an air track when it collides with a larger, 5.00kg cart at rest. After the collision, the small cart recoils at 0.830m/s.
What is the speed of the large cart after the collision? Answer in m\s.
thank you so much in advance to anyone who helps!
i think what your after is
m1*dv1 + m2*dv2 = m1*dv3 + m2*dv4
which is .1*1.9 + 5*0 = .1*.83 + 5v2
let me know if im helping

PHX
Problem 3 is .0546 m/s? eh?

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Originally Posted by Dro3
Problem 3 is .0546 m/s? eh?
correct. dro and pete  reps...i'll post a few more and give more reps out tomorrow

Banned
Originally Posted by zjp6050
correct. dro and pete  reps...i'll post a few more and give more reps out tomorrow
was i right with Q3?

Banned
Originally Posted by Dro3
Problem 3 is .0546 m/s? eh?
yeah, u're right... I forgot to account for the fact that the small cart is now moving in the opposite direction from its initial direction of travel!!
Originally Posted by PeteCraig
nah.. u did the same mistake I did. velocity of the small cart is negative in after collision..

Banned
Originally Posted by damlurker
yeah, u're right... I forgot to account for the fact that the small cart is now moving in the opposite direction from its initial direction of travel!!
lol that dus make a huge diff.
What course is this?

PHX
Originally Posted by PeteCraig
i think what your after is
m1*dv1 + m2*dv2 = m1*dv3 + m2*dv4
which is .1*1.9 + 5*0 = .1*.83 + 5v2
let me know if im helping
You didn't account for the cart recoiling backwards. It should be a negative .83.
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