Forces {SM Rev}
● Rev Q (pure tension):
● Rev Q2 (strings & pulleys):
1
,Momentum and impulse {FM1 Ch 1}
● RECALL: p(momentum)=m( mass)× v ( velocity)
● ***Impulse: change in momentum
v−u
○ For constant acceleration — F=m a=m , F t =¿ I =m v−mu
t
dv dv
○ ` variable ` — F=m a=m , ∴ F t=¿ I =m t
dt dt
●
● Vector diagram of impulse
Newton’s law of restitution - collisions
● RECALL
v 2−v1 (velocity of separation)
● ***e (coefficient of restitution[ ])=
u 1−u2 (velocity of approach)
1
○ (NOT on syllabus) e= √❑ , hb = h d where k is constant, k ≥ 1
k
1 2
mgh(GPE)= mv ( KE)→ v =√ ❑ {assume no air resistance}
2
vb
e= = √ ❑ , ¿ √ ❑
vd ❑
○ 0≤e≤1
2
, ● (perfectly inelastic when e = 0)
● https://www.geogebra.org/m/m1boNBlV
Connected particles (≥2 bodies)
●
Work, energy and power {FM1 Ch 2}
●
● RECALL:
1. At limiting equilibrium, Fmax = μR
2. When object moves at constant speed (ie a = 0), system is at dynamic
equilibrium; ONLY need F = ma when object accelerates! (otherwise Resolve)
3
● Rev Q (pure tension):
● Rev Q2 (strings & pulleys):
1
,Momentum and impulse {FM1 Ch 1}
● RECALL: p(momentum)=m( mass)× v ( velocity)
● ***Impulse: change in momentum
v−u
○ For constant acceleration — F=m a=m , F t =¿ I =m v−mu
t
dv dv
○ ` variable ` — F=m a=m , ∴ F t=¿ I =m t
dt dt
●
● Vector diagram of impulse
Newton’s law of restitution - collisions
● RECALL
v 2−v1 (velocity of separation)
● ***e (coefficient of restitution[ ])=
u 1−u2 (velocity of approach)
1
○ (NOT on syllabus) e= √❑ , hb = h d where k is constant, k ≥ 1
k
1 2
mgh(GPE)= mv ( KE)→ v =√ ❑ {assume no air resistance}
2
vb
e= = √ ❑ , ¿ √ ❑
vd ❑
○ 0≤e≤1
2
, ● (perfectly inelastic when e = 0)
● https://www.geogebra.org/m/m1boNBlV
Connected particles (≥2 bodies)
●
Work, energy and power {FM1 Ch 2}
●
● RECALL:
1. At limiting equilibrium, Fmax = μR
2. When object moves at constant speed (ie a = 0), system is at dynamic
equilibrium; ONLY need F = ma when object accelerates! (otherwise Resolve)
3
