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  1. #1
    Registered User zjp6050's Avatar
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    Physics, someone ****ing help a brother for repetitions, oops i mean reputations

    i'll post questions, if you can help that's awesome because physics just isn't my thing. i'll rep for what i have, not many reps but helping might make you feel good! thanks a lot to all who do in advance!!!

    1)

    Part A. A 1.5kg object is moving to the right with a speed of 1.1 m/s when it experiences the force shown in the figure (a). What is the object's speed after the force ends? Answer in m\s

    Part B. What is the object's direction after the force ends? (right or left?)



    Part C. A 1.5kg object is moving to the right with a speed of 1.1 m/s when it experiences the force shown in the figure (b). What is the object's speed after the force ends? Answer in m\s

    Part D. What is the object's direction after the force ends? (right or left?)
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    Registered User zjp6050's Avatar
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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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  3. #3
    Registered User zjp6050's Avatar
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    bumpis
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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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    Registered User zjp6050's Avatar
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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
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    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 = 1.65-1 = +0.65kgm/s
    Velocity = 0.65/1.5 = 0.433m/s

    d. Velocity is still positive so its still moving to the right.
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    Originally Posted by zjp6050 View Post
    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; 11-11-2008 at 10:35 PM.
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    Originally Posted by damlurker View Post
    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!
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  11. #11
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    Originally Posted by zjp6050 View Post
    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 = (0.19-0.083)/5
    V = 0.107/5

    V = 0.0214 m/s
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  12. #12
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    Originally Posted by zjp6050 View Post
    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..
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    do u still want q1 done

    cbf reading replies
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    Registered User zjp6050's Avatar
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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 View Post
    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?
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    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?
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    Originally Posted by PeteCraig View Post
    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; 11-11-2008 at 10:49 PM.
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    Originally Posted by zjp6050 View Post
    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
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    Originally Posted by damlurker View Post
    **** 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?
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    Originally Posted by damlurker View Post
    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
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    Originally Posted by zjp6050 View Post
    problem 3 was incorrect

    yes pete, you were
    **** U PHYSICS
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    all of them......







    bout tree fiddy?
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    Originally Posted by zjp6050 View Post
    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
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    Problem 3 is .0546 m/s? eh?
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    Originally Posted by Dro3 View Post
    Problem 3 is .0546 m/s? eh?
    correct. dro and pete - reps...i'll post a few more and give more reps out tomorrow
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    Originally Posted by zjp6050 View Post
    correct. dro and pete - reps...i'll post a few more and give more reps out tomorrow
    was i right with Q3?
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    Originally Posted by Dro3 View Post
    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 View Post
    was i right with Q3?
    nah.. u did the same mistake I did. velocity of the small cart is negative in after collision..
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    Originally Posted by damlurker View Post
    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?
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    Originally Posted by PeteCraig View Post
    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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