Can anyone help me with 9709 may/June/2017 paper 43 Question number 7 part (ii) I have found velocity and acceleration as $v = 1.68$ $v=1.68$ and $a = 2.83$ $a=2.83$ I don't understand how it will move further?

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Can anyone help me with 9709 may/June/2017 paper 43 Question number 7 part (ii) I have found velocity and acceleration as $v = 1.68$ $v=1.68$ and $a = 2.83$ $a=2.83$ I don't understand how it will move further?

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Answer 1 · 2018-05-16T02:42:07

Use the kinematic expression

$v^{2} - u^{2} = 2 a s$ $v^2-u^2=2as$

(Assuming your calculations are correct, Inserting given values we get

$0^{2} - {1.68}^{2} = 2 \times 2.83 s$ $0^2-1.68^2=2xx2.83s$
$\Rightarrow s = - \frac{{1.68}^{2}}{2 \times 2.83}$ $=>s=-1.68^2/(2xx2.83)$

We get a negative number. It appears your calculation of final acceleration is not correct. It should be a negative number. Perhaps next step missing.)

You need to post your work.

Answer 2 · 2018-05-16T03:36:52

Force of friction along the incline $F_{f} = μ R = μ m g cos θ$ $F_f=muR=mumgcostheta$
where $R$ $R$ is normal reaction. Taking $g=10\ ms^-2$

$F_f=0.2xx3xx10cos 30^@=3sqrt3\ N$
Once system is released from rest,
For $4\ kg$ mass
It moves down. Let its velocity as it touches ground be $v_4$ . If tension in the string is $T$ , keeping in view Law of consrvation of energy, energy equation is
$"Loss of PE"\ -"Work done against "T="Gain in KE"$
$=>4xx10xx0.5 -0.5T=1/2xx4xxv_4^2$
$=>40 -T=4v_4^2$ ......(1)
For $3kg$ mass. Let its velocity at that time be $=v_3$
$"Work done against "T="Gain in PE"+"Gain in KE"+"Work done against "F_f$
( $T$ is on both sides of smooth pulley.)
$0.5T=3xx10xx0.5sin 30^@+1/2xx3xxv_3^2+3sqrt3xx0.5$
$=>T=15+3v_3^2+3sqrt3$ ....(2)
As both masses are connected with same light inextensible string their velocities must always have equal magnitude. Let this be $=v$ . Eliminating $T$ from (1) and (2)
$40-4v^2=15+3v^2+3sqrt3$
$=>v=sqrt((25-3sqrt3)/7)=1.68\ ms^-1$
After $4\ kg$ mass touches ground $T=0$ , the net force equation for mass $3\ kg$ becomes
$"Force due to gravity along the incline"\ +"$ $F_f=F_"net"$

$–3xx10sin30 - 3sqrt3 = 3a_n$
where $a_n$ is new acceleration and $-ve$ sign is with respect to direction of motion.
$=> a_n=-5-sqrt3=-6.73\ ms^-2$
Using kinematic expression and imposing the given condition that mass $3\ kg$ comes to rest momentarily
$v^2-u^2=2as$
We get
$0^2-1.68^2=2(-6.73)s$
$=>s=1.68^2/(2xx6.73)$
$=>s=0.21\ m$
Total distance traveled by mass $3\ kg=0.5+0.21=0.71\ m$

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Can anyone help me with 9709 may/June/2017 paper 43 Question number 7 part (ii) I have found velocity and acceleration as $v = 1.68$ $v=1.68$ and $a = 2.83$ $a=2.83$ I don't understand how it will move further?

http://www.teachifyme.com/wp-content/uploads/2016/05/9709_s17_qp_43.pdf

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Can anyone help me with 9709 may/June/2017 paper 43 Question number 7 part (ii) I have found velocity and acceleration as v=1.68v=1.68v=1.68 and a=2.83a=2.83a=2.83 I don't understand how it will move further?

http://www.teachifyme.com/wp-content/uploads/2016/05/9709_s17_qp_43.pdf

2 Answers

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Can anyone help me with 9709 may/June/2017 paper 43 Question number 7 part (ii) I have found velocity and acceleration as $v = 1.68$ $v=1.68$ and $a = 2.83$ $a=2.83$ I don't understand how it will move further?