m1 (theories and conceptual mistakes)

inextensible cable assumption

Same acceleration for both objects (e.g: for a tractor and block)

just do (c ) and (d)

normally its like r = r0 + vt, but here its:
r = r0 + v(initial)t + v(new)t

OR
r = r0 + vt
where r0 ==> r(3) = r0 + v(3)

just do (b)

[a and b unrelated sums]

initially force is given to throw ball vertically upward but after that no force is given, hence it goes up due to inertia and current upward force is 0 and its decelerating

[think of after you kick a football it still keeps rising even tho there isnt anybody in the air thats pushing it or smt]

given, mass of block = 5kg, p = 14N

Find the reaction force[R] on block

R + 14sin(30) = 5g

R = 5(9.8) - 14sin(30) = 42N

the block has resistance = R and tractor resistance = 200N



find R

R = 5600N

for this type of question, friction is given as resistance so no need to bring in “μR”

p = 15×10 i = 150i , q = 20×10i = 200i

bearing of P from Q, [so draw north line from Q]

bearing = 270°

since question says magntitudes of acceleration and decelaration, we can consider deceleration without -ve sign.

try number (b), pretty tricky

(b) take Vq after collision = Vp, 1/4(2/3mu - 3u) = u/3 (because Vp= u/3)

why Vq = Vp? cuz Q. asks to find max value of m for which no further collision

[ think if you are chasing someone, both are moving at same constant v of u/3, that means you will never collide with him,only way is to have more than u/3 ]

therefore, 1/4(2/3mu - 3u) = u/3
m = 6.5 kg
to nearest integer, m = 6kg
(if m is any value > 6.5kg, they will collide which the Q. does not want, so we round down to nearest integer)

Question b)

To find the Vc right after the collision, use the conservation of momentum. The momentum of a particle is given by $p = mv$

First, find the velocities of A and B just before the collision:
For A:
$v_A = u_A + at$
$v_{A}=18-9.8(3.23)$
$v_{A}=-13.693\,\text{m/s}$

For B:
$v_B = u_B + a(t - 2)$
$v_B = 36 - 9.8(3.234 - 2)$
$v_B = 23.9068 \, \text{m/s}$

(btw Va and Vb is now Ua and Ub as we start from t=3.23 not t=0):
$m_{A}u_{A}+m_{B}u_{B}=(m_{A}+m_{B})v_{C}$
$(0.5)(-13.6932) + (0.8)(23.9068) = (0.5 + 0.8)v_C$
$-6.8466 + 19.12544 = 1.3v_C$
$12.27884 = 1.3v_C$
$v_{C}=\frac{12.27884}{1.3}=9.445\,\text{m/s}$

The answer is: 9.4 m/s

this one was tricky as normally they don’t give acceleration, so you do:
r = r0 + s
= r0 + vt

but with acceleration use: r = r0 + ut + 1/2(a)(t²)

find size of angle between direction i and resultant force, given Rf = -8i + 15j

the correct answer is the green angle, as that is size of angle theta between i and Rf

B.A = tanθ = (15/8)
θ = 61.9
size of angle = 180 - 61.9 = 118°

[red angle is wrong here as that would be angle between -i and Rf]

(also think of A, S, T ,C or cast diagram for trig, thats the way you go which is a.c.w,

hence angle is 180-61.9 not 180+61.9)

[NOTE: point of tilting means at the other support, T = 0 if its like strings or R = 0 if it’s like a seasaw]

for smallest possible value of X, particle moves down plane/point of sliding down

therefore friction opposes this motion of p, thus acts upward

• newtons 3rd law here:
  Lift pushes up on people → R (upward on people)
  People push down on lift → R (downward on lift)
  
  therefore it is T - (mass of lift)g - (force from ppl) = (mass of lift) x a
  

(this one was a bit tricky)
think about highlited region, the ball goes up until Smax, where v = 0, so it is decelerating
so we want distance from: 14.7 —> 0 and then drops down so: 0 —> 14.7

for a threaded ring/bead onto a fixed rough vertical rod: the reaction, R is to left or right not up
(depending on tension direction),

(The reaction R is perpendicular to the rod )

distance closest together OR min distance = 0.707m

( a really nice equilibrium question)

when in equilibrium, T - mg = 0, => T = mg

for (b) you cud do just use cos rule to find angle, much easier

NOTE:
since a string is used and its threaded, there is no R.

Find the possible values of K
(5marks)

main concept here:
direction of P after collision is not given
So P it can go → or ← after collsion, hence 2 values of K will be found

just concept →

For vector sums, keep in terms of i and j and at last convert to magnitude if needed
(like if they ask for speed then convert to magnitude)

for Suvat sums, especially the tricky ones, keep in terms of g and if needed x 9.8
you might lose marks due to inaccuracy if u input 9.8 early on

State how you have used the fact that the beam is modelled as a rod

(1mark)

- The beam is rigid and remains straight