M1 Help - 3 problems

Hi

Need help in these mechanics problems.

They are from review questions.
21. Two particles A an B of masses 0.4 kg and 0.3 kg respectively are connected by a light inextensible string which passes over a smooth fixed pulley. The particles are released from rest with the string taut and hanging parts vertical.
I’ve got the answer to parts a )and b.) I don’t know how to get the third part.

The particles continue to move in this system until the instant when they have each acquired speed 3.5m/s. At this instant both A and B are 1.4 m above horizontal ground and the string is cut.
Given that each particle now moves freely under gravity, find the difference in the times, measured from the instant when the string is cut, for A and B to reach the ground.

The other question is no. 29( d )where two particles, A and B of mass 0.2 kg and o.5 kg are connected , smooth pulley. The particles are released from rest with the string taut and the hanging parts vertical. Calculate the total time from the instant when the particles were first released until the instant when A first comes to instantaneous rest. (Assume that A does not reach the pulley.

And the last question is 36 (c ) A force R acts on a particle, where R = (7i + 16j)N . The force R is the resultant of two forces P and Q. The line of action of P is parallel to the vector I + 4j) and the line of action of Q is parallel to the vector (I + j). Determine the forces P and Q expressing each in terms of I and j

Thanx in advance, any help much appreciated.

Reply 1

C4>O7

5

Resolving vertically for A:
0.4g-T=0.4a
" " for B:
T-0.3g=0.3a

(1)+(2) => 0.1g=0.7a
=> a=1/7gms^-2

Sub. in (1) for a=> T=12/35gN

For A:
s=ut+½at²
-1.4 = -3.5t-4.9t²
(t+1)(4.9t-1.4)=0
t=(1.4/4.9)s for A to reach ground

For B:
s=ut+½at²
-1.4=3.5t-4.9t²
(½t-½)(9.8t+2.8)=0
t=1s for B to reach ground

t₁-t₂ = 0.714s

Reply 2

Chewwy

17

shivmani

And the last question is 36 (c ) A force R acts on a particle, where R = (7i + 16j)N . The force R is the resultant of two forces P and Q. The line of action of P is parallel to the vector I + 4j) and the line of action of Q is parallel to the vector (I + j). Determine the forces P and Q expressing each in terms of I and j

i havent done M1 yet, i haven't done any a-level modules yet, but i think it goes like this:

p -> (i + 4j)
q -> (i + j)

x(i+4j) + y(i+j) = (7i+16j)

x + y = 7
4x + y = 16

...

x = 3
y = 4

so:

p = 3(i+4j) = (3i + 12j)
q = 4(i+j) = (4i + 4j)

-----------

now, i could be barking completely up the wrong tree, so someone confirm this please...

no that's right

C4>O7

no that's right

hurrah!

Reply 5

shivmani

OP

C4>O7

Resolving vertically for A:
0.4g-T=0.4a
" " for B:
T-0.3g=0.3a

(1)+(2) => 0.1g=0.7a
=> a=1/7gms^-2

Sub. in (1) for a=> T=12/35gN

For A:
s=ut+½at²
-1.4 = -3.5t-4.9t²
(t+1)(4.9t-1.4)=0
t=(1.4/4.9)s for A to reach ground

For B:
s=ut+½at²
-1.4=3.5t-4.9t²
(½t-½)(9.8t+2.8)=0
t=1s for B to reach ground

t₁-t₂ = 0.714s

Hi!

I don't understand why s and ut and at have been taken as negative for A whereas s and at for B is negative while ut is positive.

Also, could anyone please try 29d? Thanks a ton!

Reply 6

Fermat

8

For A:
s is 1.4 m down :. -ve
a = 9.8 m/s² down :. -ve
u = 3.5 m/s down :. -ve

For B:
s is 1.4 m down :. -ve
a = 9.8 m/s² down :. -ve
u = 3.5 m/s up :. +ve

Reply 7

Fermat

8

shivmani

Hi!

I don't understand why s and ut and at have been taken as negative for A whereas s and at for B is negative while ut is positive.

Also, could anyone please try 29d? Thanks a ton!

For Q. 29d,

What is it that makes A come to rest ?

Is particle B halted ?

How ?

Is the string cut ?

Reply 8

shivmani

OP

sorry, should have written this before
part a) when the particles released from rest, i got the initial acceleration as 4.2m/s^2.
At the instant when A and B are moving with speed 6.3 m/s B strikes an inelastic horizontal floor from which it does not rebound. Calculate the impulse exerted on the floor by B (I got that as 3.15 Ns) and the total time from the instant when the particles were first released until the instant when A first comes to instantaneous rest. Need help on the time bit.
thanks

Reply 9

Fermat

8

29/
a)
F = ma
(0.5 - 0.2)g = (0.5 + 0.2)a
a = (0.3/0.7)*9.8
a = 4.2 m/s²
=========

b)
I = change in momemtum
I = 0.5*6.3
I = 3.15 Ns
========

c)
let t1 be the time taken to reach speed of 6.3 m/s

v = u + at - SUVAT eqn
6.3 = 0 + 4.2t1
t1 = 6.3/4.2
t1 = 1.5 s
=======

let t2 be the additional time required to come to rest.

v = u + at - SUVAT eqn
0 = 6.3 - gt2
t2 = 6.3/9.8
t2 = 0.643 s
=========

Total time is t = t1 + t2
t = 1.5 + 0.643
t = 2.143 s
========

btw, do you take g as 9.8 or 10 in M1 ?