Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Can you help me with the following questions, thank you.

    1) A particle X moving along a straight line with constant speed 4ms-1 passes through a fixed point 0. Two seconds later another particle Y moving along the same straight line and in the same direction passes through 0 with speed 6ms-1. Given that Y is subject to a constant deceleration of magnitude 2ms-2,

    a) state the velocity and displacement of each particle t seconds after Y passed through O.
    B) find the shortest distance between the particles after they have both passed though 0.
    c) find the value of t when the distance between the particles has increased to 23m.

    2) A particle P moves from rest at a point 0 at time t=0 seconds, along a straight line. At any subsequent time t seconds the acceleration of P is proportional to 7-tsquared ms-2 and the displacement of P from 0 is s metres. The spped of P is 6ms-1 when t=3.

    a) Show that s = 1/24tsquared(42 - tsquared)
    b) find the total distance in metres that P moves before returning to 0.
    Offline

    2
    ReputationRep:
    (Original post by hussein)
    Can you help me with the following questions, thank you.

    1) A particle X moving along a straight line with constant speed 4ms-1 passes through a fixed point 0. Two seconds later another particle Y moving along the same straight line and in the same direction passes through 0 with speed 6ms-1. Given that Y is subject to a constant deceleration of magnitude 2ms-2,

    a) state the velocity and displacement of each particle t seconds after Y passed through O.
    B) find the shortest distance between the particles after they have both passed though 0.
    c) find the value of t when the distance between the particles has increased to 23m.
    1)
    a)
    Vx = const = 4 m/s
    Sx = Vx(t+2) = 4(t+2)

    Vy = 6 - 2t
    Sy = Vy*t = 6t -t²

    b)
    S = Sx - Sy
    S = 4(t+2) - 6t + t²
    S = 8 - 2t + t²

    Now differentiate S and equate it to zero to get a minimum.

    dS/dt = -2 + 2t = 0 @ t=1

    Smin = 8 - 2 + 1 = 7m

    c)
    S = 23
    23 = 8 - 2t + t²
    t² - 2t - 15 = 0
    (t - 5)(t + 3) = 0
    t = 5 or t = -3

    Choose t = 5 since distance is increasing to 25m rather than decreasing from 23 m (which is when the t=3 value would apply)
    Offline

    13
    ReputationRep:
    Here's a question. I won't bother making a new topic...

    M2 pg 107 Q8.

    A sphere of mass m1 moving with speed u1 collides directly with a similar sphere of mass m2 moving with speed u2 in the same direction (u1>u2). The coefficient of restitution between the two spheres is e. Show that the loss of kinetic energy E due to the collision satisfies the equation:

    2 (m1+m2) E = m1 m2 (u1-u2)^2 (1-e^2)

    (BTW m1 means m subscript 1. Same for m2 u1 u2).
    Offline

    2
    ReputationRep:
    (Original post by hussein)
    Can you help me with the following questions, thank you.

    2) A particle P moves from rest at a point 0 at time t=0 seconds, along a straight line. At any subsequent time t seconds the acceleration of P is proportional to 7-tsquared ms-2 and the displacement of P from 0 is s metres. The spped of P is 6ms-1 when t=3.

    a) Show that s = 1/24tsquared(42 - tsquared)
    b) find the total distance in metres that P moves before returning to 0.
    2)
    a)

    a = k(7-t²)

    a = d²s/dt² = dv/dt = k(7-t²)

    integrating the dv/dt = k(7-t²),

    v = k(7t - t³/3) + C

    at t = 0, v = 0 => C = 0

    v = k(7t - t³/3)

    at t = 3, v = 6,

    6 = k(21 - 9)
    k = ½
    v = ½(7t - t³/3)
    ===========

    ds/dt = v = ½(7t - t³/3)

    integrating ds/dt = ½(7t - t³/3),

    s = ½(7t²/2 - t^4/12) + D

    at t = 0, s = 0, => D = 0

    s = ½(7t²/2 - t^4/12)
    s = t²(42 - t²)/24
    ===========

    b)
    To find max s, differentiate s and equate it to zero.

    ds/dt = (84t - 4t³)/24 = 0
    t(21 - t²) = 0

    s is a max at t = √21

    s_max = 21(42 - 21)/24
    s_max = 18.375

    P will move 18.375m away from O and then move another 18.375 m back to O. So total distance travelled is,

    s_total = 36.75 m
    ===========
    Offline

    2
    ReputationRep:
    (Original post by SsEe)
    Here's a question. I won't bother making a new topic...

    M2 pg 107 Q8.

    A sphere of mass m1 moving with speed u1 collides directly with a similar sphere of mass m2 moving with speed u2 in the same direction (u1>u2). The coefficient of restitution between the two spheres is e. Show that the loss of kinetic energy E due to the collision satisfies the equation:

    2 (m1+m2) E = m1 m2 (u1-u2)^2 (1-e^2)

    (BTW m1 means m subscript 1. Same for m2 u1 u2).
    This gives me a chance to practice on my latex - sure do need it !
    Attached Images
  1. File Type: pdf collision.pdf (36.3 KB, 942 views)
 
 
 
Turn on thread page Beta
Updated: June 22, 2004
Poll
Do you think parents should charge rent?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.