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Laws of Motion

Pa ge 1 0 9
Quest ion 5.1:
Give t he m agnit ude and direct ion of t he net force act ing on
( a) a drop of rain falling dow n w it h a const ant speed,
( b) a cork of m ass 10 g float ing on wat er,
( c) a k it e sk illfully held st at ionary in t he sky,
( d) a car m oving wit h a const ant v elocit y of 30 km / h on a rough road,
( e) a high- speed elect ron in space far from all m at erial obj ect s, and free of elect r ic
and m agnet ic fields.
ANS:
( a ) Zer o net force
The rain dr op is falling wit h a const ant speed. Hence, it accelerat ion is zero. As per
Newt on’s second law of m ot ion, t he net force act ing on t he rain drop is zero.
( b) Zero net force
The w eight of t he cork is act ing downward. I t is balanced by t he buoyant force
ex ert ed by t he wat er in t he upwar d direct ion. Hence, no net force is act ing on t he
float ing cork .
( c) Zero net force
The k it e is st at ionary in t he sk y, i.e., it is not m ov ing at all. Hence, as per Newt on’s
first law of m ot ion, no net force is act ing on t he kit e.
( d) Zero net force
The car is m ov ing on a rough road w it h a const ant v elocit y. Hence, it s accelerat ion is
zero. As per Newt on’s second law of m ot ion, no net force is act ing on t he car.
( e ) Zero net force

Page 1 of 49

Laws of Motion
The high speed elect ron is free from t he influ ence of all fields. Hence, no net force is
act ing on t he elect ron.
Quest ion 5.2:
A pebble of m ass 0.05 k g is t hr own v ert ically upwards. Give t he direct ion and
m agnit ude of t he net force on t he pebble,
( a) during it s upwar d m ot ion,
( b) during it s dow nwar d m ot ion,
( c) at t he highest point where it is m om ent arily at rest . Do y our answers change if
t he pebble was t hrown at an angle of 45° wit h t he horizont al direct ion?
I gnor e air resist ance.
ANS:
0.5 N, in vert ically dow nward direct ion, in all cases
Accelerat ion due t o grav it y, irrespect ive of t he direct ion of m ot ion of an obj ect ,
alway s act s dow nward. The grav it at ional force is t he only force t hat act s on t he
pebble in all t hree cases. I t s m agnit ude is giv en by Newt on’s second law of m ot ion
as:
F= m × a
Where,
F = Net force
m = Mass of t he pebble = 0.05 k g
a = g = 10 m / s2
∴ F = 0.05 × 10 = 0.5 N
The net force on t he pebble in all t hree cases is 0.5 N and t his force act s in t he
downward direct ion.
I f t he pebble is t hrow n at an angle of 45° wit h t he horizont al, it w ill have bot h t he
hor izont al and v ert ical com ponent s of v elocit y. At t he highest point , only t he v ert ical
com ponent of v elocit y becom es zero. However, t he pebble w ill hav e t he horizont al
com ponent of v elocit y t hroughout it s m ot ion. This com ponent of velocit y pr oduces no
effect on t he net force act ing on t he pebble.

Page 2 of 49

Laws of Motion
Quest ion 5.3:
Give t he m agnit ude and direct ion of t he net force act ing on a st one of m ass 0.1 k g,
( a) j ust aft er it is dropped from t he w indow of a st at ionary t rain,
( b) j ust aft er it is dropped from t he w indow of a t rain running at a const ant velocit y
of 36 km / h,
( c) j ust aft er it is dropped from t he w indow of a t rain accelerat ing w it h 1 m s –2 ,
( d) ly ing on t he floor of a t rain w hich is accelerat ing w it h 1 m s –2 , t he st one being at
rest relat iv e t o t he t rain. Neglect air resist ance t hroughout .
ANS:
( a ) 1 N; vert ically dow nward
Mass of t he st one, m = 0.1 k g
Accelerat ion of t he st one, a = g = 10 m / s2
As per Newt on’s second law of m ot ion, t he net force act ing on t he st one,
F = ma = mg
= 0.1 × 10 = 1 N
Accelerat ion due t o grav it y alw ays act s in t he downwar d direct ion.
( b) 1 N; v ert ically dow nward
The t rain is m ov ing w it h a const ant v elocit y . Hence, it s accelerat ion is zero in t he
direct ion of it s m ot ion, i.e., in t he horizont al dir ect ion. Hence, no force is act ing on
t he st one in t he horizont al direct ion.
The net force act ing on t he st one is because of accelerat ion due t o grav it y and it
alway s act s vert ically downwar d. The m agnit ude of t his force is 1 N.
( c) 1 N; vert ically dow nward
I t is giv en t hat t he t rain is accelerat ing at t he rat e of 1 m / s 2 .
Therefore, t he net force act ing on t he st one, F’ = m a = 0.1 × 1 = 0.1 N
This force is act ing in t he hor izont al direct ion. Now, when t he st one is dropped, t he
hor izont al force F,’ st ops act ing on t he st one. This is because of t he fact t hat t he

Page 3 of 49

Laws of Motion
force act ing on a body at an inst ant depends on t he sit uat ion at t hat inst ant and not
on ear lier sit uat ions.
Therefore, t he net force act ing on t he st one is giv en only by accelerat ion due t o
grav it y.
F = mg = 1 N
This force act s vert ically dow nward.
( d) 0.1 N; in t he direct ion of m ot ion of t he t rain
The w eight of t he st one is balanced by t he norm al react ion of t he floor. The only
accelerat ion is prov ided by t he hor izont al m ot ion of t he t rain.
Accelerat ion of t he t rain, a = 0.1 m / s2
The net force act ing on t he st one w ill be in t he direct ion of m ot ion of t he t rain. I t s
m agnit ude is given by:
F = ma
= 0.1 × 1 = 0.1 N

PAGE 1 1 0
Quest ion 5.4:
One end of a st ring of lengt h l is connect ed t o a part icle of m ass m and t he ot her t o
a sm all peg on a sm oot h hor izont al t able. I f t he part icle m oves in a circle w it h speed
v t he net force on t he part icle ( direct ed t owar ds t he cent re) is:

( i) T, ( ii)

, ( iii)

, ( iv) 0

T is t he t ension in t he st ring. [ Choose t he correct alt ernat iv e] .

Page 4 of 49

Laws of Motion
ANS:
An sw e r : ( i)
When a part icle connect ed t o a st r ing rev olv es in a circular pat h around a cent r e, t he
cent ripet al force is provided by t he t ension produced in t he st ring. Hence, in t he
giv en case, t he net force on t he part icle is t he t ension T, i.e.,

F= T=
Where F is t he net force act ing on t he part icle.

Quest ion 5.5:
A const ant ret ar ding force of 50 N is applied t o a body of m ass 20 k g m ov ing init ially
wit h a speed of 15 m s –1 . How long does t he body t ake t o st op?
ANS:
Ret arding force, F = –50 N
Mass of t he body, m = 20 k g
I nit ial v elocit y of t he body , u = 15 m / s
Final v elocit y of t he body, v = 0
Using Newt on’s second law of m ot ion, t he accelerat ion ( a) produced in t he body can
be calculat ed as:
F = ma
–50 = 20 × a

Page 5 of 49

Laws of Motion
Using t he first equat ion of m ot ion, t he t im e ( t ) t aken by t he body t o com e t o rest can
be calculat ed as:
v = u + at

= 6 s

Quest ion 5.6:
A const ant force act ing on a body of m ass 3.0 k g changes it s speed from 2.0 m s –1 t o
3.5 m s–1 in 25 s. The direct ion of t he m ot ion of t he body rem ains unchanged. What
is t he m agnit ude and direct ion of t he force?
ANS:
0.18 N; in t he direct ion of m ot ion of t he body
Mass of t he body, m = 3 k g
I nit ial speed of t he body , u = 2 m / s
Final speed of t he body, v = 3.5 m / s
Tim e, t = 25 s
Using t he first equat ion of m ot ion, t he accelerat ion ( a) pr oduced in t he body can be
calculat ed as:
v = u + at

As per Newt on’s second law of m ot ion, force is giv en as:
F = ma
= 3 × 0.06 = 0.18 N

Page 6 of 49

Laws of Motion
Since t he applicat ion of force does not change t he dir ect ion of t he body , t he net force
act ing on t he body is in t he dir ect ion of it s m ot ion.

Quest ion 5.7:
A body of m ass 5 k g is act ed upon by t w o perpendicular forces 8 N and 6 N. Giv e t he
m agnit ude and direct ion of t he accelerat ion of t he body .
ANS:
2 m / s2 , at an angle of 37° w it h a force of 8 N
Mass of t he body, m = 5 k g
The giv en sit uat ion can be represent ed as follows:

The result ant of t w o for ces is giv en as:

θ is t he angle m ade by R wit h t he force of 8 N

The negat iv e sign indicat es t hat θ is in t he clock wise direct ion w it h respect t o t he
force of m agnit ude 8 N.

Page 7 of 49

Laws of Motion
As per Newt on’s second law of m ot ion, t he accelerat ion ( a) of t he body is given as:
F = ma

Quest ion 5.8:
The driver of a t hr ee- w heeler m ov ing w it h a speed of 36 km / h sees a child st anding
in t he m iddle of t he road and br ings his vehicle t o rest in 4.0 s j ust in t im e t o save
t he child. What is t he average r et arding force on t he v ehicle? The m ass of t he t hreewheeler is 400 k g and t he m ass of t he driv er is 65 k g.
ANS:
I nit ial speed of t he t hree- wheeler, u = 36 km / h
Final speed of t he t hree- wheeler, v = 10 m / s
Tim e, t = 4 s
Mass of t he t hree- wheeler, m = 400 k g
Mass of t he driver, m ’ = 65 k g
Tot al m ass of t he syst em , M = 400 + 65 = 465 k g
Using t he first law of m ot ion, t he accelerat ion ( a) of t he t hr ee- w heeler can be
calculat ed as:
v = u + at

The negat iv e sign indicat es t hat t he velocit y of t he t hree- w heeler is decreasing w it h
t im e.
Using Newt on’s second law of m ot ion, t he net force act ing on t he t hree- wheeler can
be calculat ed as:
F = Ma
= 465 × ( –2.5) = –1162.5 N

Page 8 of 49

Laws of Motion
The negat iv e sign indicat es t hat t he force is act ing against t he dir ect ion of m ot ion of
t he t hree- w heeler.

Quest ion 5.9:
A rock et w it h a lift - off m ass 20,000 k g is blast ed upwards w it h an init ial accelerat ion
of 5.0 m s–2 . Calculat e t he init ial t hrust ( force) of t he blast .
ANS:
Mass of t he rock et , m = 20,000 k g
I nit ial accelerat ion, a = 5 m / s2
Accelerat ion due t o grav it y, g = 10 m / s2
Using Newt on’s second law of m ot ion, t he net force ( t hr ust ) act ing on t he rocket is
giv en by t he relat ion:
F – mg = ma
F = m ( g + a)
= 20000 × ( 10 + 5)
= 20000 × 15 = 3 × 10 5 N

Quest ion 5.10:
A body of m ass 0.40 k g m ov ing init ially w it h a const ant speed of 10 m s –1 t o t he
nort h is subj ect t o a const ant force of 8.0 N direct ed t owards t he sout h for 30 s.
Take t he inst ant t he for ce is applied t o be t = 0, t he posit ion of t he body at t hat t im e
t o be x = 0, and predict it s posit ion at t = –5 s, 25 s, 100 s.
ANS:
Mass of t he body, m = 0.40 k g
I nit ial speed of t he body , u = 10 m / s due nort h
Force act ing on t he body, F = –8.0 N

Accelerat ion produced in t he body ,

Page 9 of 49


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