Laws of Motion

Question 513
CBSEENPH11020529

A particle of mass m is driven by a machine that delivers a constant power K watts. If the particle starts from rest, the force on the particle at time t is

  • square root of mk over 2 end root t to the power of negative 1 divided by 2 end exponent
  • square root of mk space t to the power of negative 1 divided by 2 end exponent
  • square root of 2 mk end root space t to the power of negative 1 divided by 2 end exponent
  • 1 half square root of m t end root to the power of negative 1 divided by 2 end exponent

Solution

A.

square root of mk over 2 end root t to the power of negative 1 divided by 2 end exponent

As the machine delivers a constant power
So F, v =constant = k (watts)
rightwards double arrow space straight m dv over dt. straight v space equals straight k
rightwards double arrow integral vdv space equals straight k over straight m integral dt
rightwards double arrow straight v squared over 2 equals straight k over straight m straight t space space
rightwards double arrow space straight v equals space square root of fraction numerator 2 straight k over denominator straight m end fraction end root straight t
Now comma space force space on space the space particles space is space given space by
straight F equals space straight m fraction numerator d straight v over denominator d straight t end fraction space equals straight m straight d over dt open parentheses fraction numerator 2 kt over denominator straight m end fraction close parentheses to the power of 1 divided by 2 end exponent
equals space square root of 2 km end root space open parentheses 1 half straight t to the power of negative 1 half end exponent close parentheses
equals space square root of mk over 2 end root. straight t to the power of fraction numerator negative 1 over denominator 2 end fraction end exponent

Question 514
CBSEENPH11020538

Three blocks A, B, and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a 14 N is applied to the 4 kg block, then the contact force between A and B is

  • 2 N

  • 6 N

  • 8 N 

  • 18 N 

Solution

B.

6 N

Given, mA = 4 kg
mB = 2 kg 
=> mC =1 kg

So total mass (M)  = 4+2+1 = 7 kg
Now, F = Ma 
14 = 7a
a=2 m/s2

F-F' = 4a
F' = 14-4x2
F' = 6N

Question 515
CBSEENPH11020539

A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of the table and from its other end, another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is straight mu subscript straight k. When the block A is sliding on the table, the tension in the string is

  • fraction numerator left parenthesis straight m subscript 2 plus straight mu subscript straight k straight m subscript 1 right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction
  • fraction numerator left parenthesis straight m subscript 2 minus straight mu subscript straight k straight m subscript 1 right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction
  • fraction numerator straight m subscript 1 straight m subscript 2 left parenthesis 1 plus straight mu subscript straight k right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction
  • fraction numerator straight m subscript 1 straight m subscript 2 left parenthesis 1 minus straight mu subscript straight k right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction

Solution

C.

fraction numerator straight m subscript 1 straight m subscript 2 left parenthesis 1 plus straight mu subscript straight k right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction

FBD of block A,

T -m1afk ..... (i)
FBD  of block B

m2g -T = m2a ... (ii)
Adding Eqs. (i) and (ii), we get
m2g-m1a = m2a +fkrightwards double arrow space straight m subscript 2 straight g minus straight m subscript 1 straight a space equals space straight m subscript 2 straight a space plus straight f subscript straight k
rightwards double arrow space straight a equals space fraction numerator left parenthesis straight m subscript 2 minus straight u subscript straight k straight m subscript 1 right parenthesis straight g over denominator straight m subscript 1 plus straight m subscript 2 end fraction
from space equation space left parenthesis ii right parenthesis
straight T equals space straight m subscript 2 minus left parenthesis straight g minus straight a right parenthesis
equals space straight m subscript 2 open square brackets 1 minus fraction numerator left parenthesis straight m subscript 2 minus straight mu subscript straight k straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight g
straight T equals space fraction numerator straight m subscript 1 straight m subscript 2 space left parenthesis 1 plus straight mu subscript straight k right parenthesis over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight g
Question 516
CBSEENPH11020548

A stone is dropped from a height h. It hits the ground with a certain momentum p. If the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by

  • 68%

  • 41%

  • 200%

  • 100%

Solution

B.

41%

velocity space straight v space equals space square root of 2 gh end root space space space space space space space.... left parenthesis straight i right parenthesis
and space momentum space straight p space equals space mv space space space space... space left parenthesis ii right parenthesis
From space left parenthesis Eqs. right parenthesis space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space have
space straight p proportional to square root of straight h
Here comma
straight p subscript 2 over straight p subscript 1 space equals space square root of straight h subscript 2 over straight h subscript 1 end root
therefore space fraction numerator straight p subscript 2 over denominator straight p subscript 1 space end fraction space equals square root of fraction numerator 2 straight h over denominator straight h end fraction space end root space equals square root of 2

straight p subscript 2 space equals 1.414 straight p subscript 1
percent sign change space equals space fraction numerator straight p subscript 2 minus straight p subscript 1 over denominator straight p subscript 1 end fraction space straight x space 100 space equals space 41 percent sign

NCERT Book Store

NCERT Sample Papers

Entrance Exams Preparation