M1 Mechanics questions!

EX 3C, M1

7) A boy is walking at a constant speed of 1.8m/s along a straight road. He passes a telephone booth where his sister is making a telephone call. His sister takes 30 seconds to complete the call and then sets off in pursuit of the boy. She accelerates uniformly from rest at 3m/s until she is running at a speed of 9m/s. She maintains this constant speed until she reaches her brother.

b) calculate the time taken by the girl to reach her brother.

EX 3D,

4) A girl throws a ball vertically upwards with speed 8m/s from a window which is 6 metres above horizontal ground.
1.5 seconds later she drops a second ball from rest out of the same window.
b) find the distance below the window of the point where the balls meet.

8) A boy drops a stone from rest from the top of a tower which is 25 m high. In an initial model, the stone is assumed to be a particle and air resistance is negligible.
In a refined model air resistance is not negligible and it takes 0.2 seconds longer for the bll to reach the ground.

calculate the revised acceleration of the stone

Pls explain these questions step by step,
thanks a lot

Reply 1

m00c0w06

7b.

s = ut so s = 1.8 * 30 = 54m

Now we need to work out how long it takes and how far the girl travels until she reaches 9m/s.

u = 0
v = 9
a = 3
t = ?
s = ?

v = u + at
9 = 0 + 3t
t = 3 seconds
s = 1/2(u+v)t = 1/2 * 9 * 3 = 13.5m

In those 3 seconds the boy has travelled 3 * 1.8 = 5.4m

54 + 5.4 - 13.5 = 45.9m

So using s = ut + 1/2at^2...

For the boy:

s = 1.8t

For the girl:

s + 45.9 = 9t
s = 9t - 45.9

Solving simultaneously:

1.8t = 9t - 45.9
7.2t = 45.9
t = 6.375s

Reply 2

Mohit_C

Rite, i dont get why you do 54+5.4 then take away 13.5m??

Reply 3

x kat x

now i know why i had a 'u' in that topic. i didn't get a word of that!

Reply 4

Mohit_C

The answer you got is wrong, when i checked it with that of the book.

Reply 5

*bobo*

your answers are wrong.

firstly you must consider how far the boy has travelled before his sister even begins. Hes travelling at 1.8m/s for 30 secs. so hes travelled 54m.

so his equation for how far hes travelled is: 54+ 1.8 t (t being the time since his sister began)

for the sister first you must consider how far shes travelled during accelerating to her maximum speed.
use a= (v-u)/t
a= 3 , u=0 and v=9
so t=3secs

and using v^2=u^2+2as
s= 13.5

so the equation for how far shes travelled is:
13.5 + (t-3)9 (because shes already spent 3 secs travelling to achieve 13.5m, and then tavels at 9m/s for the remainding time)

so:
54 + 1.8t= 13.5 +(t-3)9

and you deduce that t=9.375 secs

Reply 6

Mohit_C

Ok, the actual time is 39.4 seconds....how is that??

Reply 7

*bobo*

Mohit_C

Ok, the actual time is 39.4 seconds....how is that??

ahh, i thought they wanted the time from when the girl started. just add 30 to my original answer 9.375secs. you add 30 because this is how long it took before the girl started

Reply 8

shivmani2

Ex 3D no. 8

a = 9.8, s =9.8, u =0, V=? , t =?

use V^2=u^s+2as to find v
v=22.14
t=(v-u)/t = 2.26s

b) u=0, s=25 a =? t =2.26+0.2

s= ut+o.5at^2

a=8.26

Reply 9

Mohit_C

Ok guys, thanks a lot. Only need help with that window thing.

Reply 10

Mohit_C

Got another two questions:

A body of mass 2 kg is held in limiting equilibrium on a rough plane inclined at 20º to the horizontal by a horizontal force X. The coefficient of friction between the body and the plane is 0.2. Modelling the body as a particle find X when the body is on the point of slipping

a) up the plane
b) down the plane

Thanks a lot.

Reply 11

mizfissy815

EX 3D,

4) A girl throws a ball vertically upwards with speed 8m/s from a window which is 6 metres above horizontal ground.
1.5 seconds later she drops a second ball from rest out of the same window.
b) find the distance below the window of the point where the balls meet.

(a) +ve ↑
u = 8 ms^–1   a = –9.8 ms^–2   v = 0 ms^–1   s = h m
v² = u² + 2as
0 = 82 – 2 × 9.8h
h = 64
19.6
= 3.265
Greatest height above the ground = (6 + 3.27) m = 9.27 m

(b)

Let the balls meet x metres below the window, T seconds after the first ball was thrown.
For the first ball (taking positive direction downwards):
u = –8 ms^–1 a = 9.8 ms^–2   s = x m   t = T s
s = ut + 1/2 at²
x = –8T + 1/2 × 9.8T².... equation no. 1

For the second ball:
u = 0 ms^–1 a = 9.8 ms^–2 s = x m t = (T – 1.5) s
x = 1/2 × 9.8(T – 1.5)² ....equation no. 2

Eliminating x between 1 and 2:
4.9(T – 1.5)² = –8T + 4.9T²
4.9(T² – 3T + 2.25) = –8T + 4.9T²
4.9T² – 14.7T + 4.9 × 2.25 = –8T + 4.9T²
14.7T – 8T = 4.9 × 2.25
T = (4.9×2.25) / 6.7 = 1.645

Using 2: x = 1/2 × 9.8(T – 1.5)² = 0.104
The balls meet 0.104 m below the window.

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Reply 12

mizfissy815

A body of mass 2 kg is held in limiting equilibrium on a rough plane inclined at 20º to the horizontal by a horizontal force X. The coefficient of friction between the body and the plane is 0.2. Modelling the body as a particle find X when the body is on the point of slipping

a) up the plane

(a) Resolving vertically,
R cos 20° – 0.2R sin 20° – 2g = 0
R(cos 20° – 0.2 sin 20°) = 19.6
R = 22.50
Resolving horizontally,
X – 0.2R cos 20° – R cos 70° = 0
X = R(0.2 cos 20° + cos 70°)
= 22.50(0.2 cos 20° + cos 70°)
= 11.9

(b) Resolving vertically,
R cos 20° + 0.2R sin 20° – 2g = 0
R(cos 20° + 0.2 sin 20°) = 2g
R = 19.44
Resolving horizontally,
X – R sin 30° + 0.2R cos 20° = 0
X = R(sin 20° – 0.2 cos 20°)
= 19.44(sin 20° – 0.2 cos 20°)
= 3.00
Force required is 3.00 N

or...
1) R= mgcos20 +Xsin20
2) Xcos20 + µR = mgsin20
so...sub 1) into 2)

Xcos20= 19.62sin20 - [0.2 x (19.62cos20+Xsin20)]
Xcos20= 6.71-3.924cos20 - 0.2Xsin20
0.2Xsin20 + Xcos20 = 3.023081368
1.0081 X=3.023081368
X= 2.998≈ 3.00N

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Reply 13

Mohit_C

Thanks a lot, it's helped me quite a lot!!

Reply 14

Mohit_C

Ok, the question of the body slipping down the slope/up the slope is repeated again, but this time the force is 45º to the horizontal. For some reason, i kept on trying it but i couldn't get the answer. Pls help.

Reply 15

Mohit_C

Got another three questions here:

1. A stone slides in a straight line across a frozen pond. Given that the initial speed of the stone is 5 M/s and that it slides 20 metres before coming to rest, calculate the coefficient of friction between the stone and the surface of the frozen pond. [ EX 5A]

2. A boy is tobogganing down a smooth slope inclined at 20º to the horizontal. Find his acceleration, assuming resistances can be neglected. [EX 5B]

3. A particle of mass 2 kg rests in limiting equilibrium on a plane inclined at 25º to the horizontal. The angle of inclination is decreased to 20º and a force of magnitude 20 N is applied up a line of greatest slope. Find the particle's acceleration [done this]. When the particle has been moving for 2 seconds the force is removed. Determine the further distance the particle will move up the plane.

Thanks in advance.

Reply 16

manan23

Original post by shivmani2

Ex 3D no. 8

a = 9.8, s =9.8, u =0, V=? , t =?

use V^2=u^s+2as to find v
v=22.14
t=(v-u)/t = 2.26s

b) u=0, s=25 a =? t =2.26+0.2

s= ut+o.5at^2

a=8.26

why does s=9.8???? shouldnt it be 25?