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# Mechanics 2 Help watch

1. 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.
2. (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)
3. 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).
4. (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
===========
5. (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
6. collision.pdf (36.3 KB, 942 views)

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