Laws of Motion

Question 541
CBSEENPH11020753

A shell of mass 200 g is ejected from a gun of mass 4 g by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is

  • 100 ms-1

  • 80 ms-1

  • 40 ms-1

  • 120 ms-1

Solution

A.

100 ms-1

In the given problem conservation of linear momentum and energy hold good.
Conservation of momentum yields.
m1v1 + m2v2 = 0
4v1 + 0.2 v2  = 0
Conservation of energy yields

1 half straight m subscript 1 straight v subscript 1 squared space plus 1 half space straight m subscript 2 straight v subscript 2 squared space equals space 1050 space space space... left parenthesis straight i right parenthesis
or
1 half space straight x 4 straight v subscript 1 squared space plus space 1 half 0.1 straight v subscript 2 squared space equals space 1050 space space space.... space left parenthesis ii right parenthesis

solving space eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space have
straight v subscript 1 space equals space 100 space straight m divided by straight s

Question 542
CBSEENPH11020759

Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v m/s will be

  • Mv newton

  • 2 Mv newton

  • Mv/2 Newton

  • zero

Solution

A.

Mv newton

Force required,
straight F space equals space fraction numerator straight d left parenthesis mv right parenthesis over denominator dt end fraction
equals space straight v open parentheses dm over dt close parentheses space equals space vM

as velocity v is constant, hence
F= Mv newton

Question 543
CBSEENPH11020772

A block B is pushed momentarily along a horizontal surface with an initial velocity v. if μ is the coefficient of sliding friction between B  and the surface, block B will come to rest after a time:

  • v/gμ

  • gμ/v

  • g/v

  • v/g

Solution

A.

v/gμ

Block B will come to rest, if force applied to it will vanish due to frictional force acting between block B and surface, ie,
force applied = frictional force
i.e., μmg = ma
μmg = m (v/t)
t =v/μg

Question 544
CBSEENPH11020795

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking =10 m/s2, work done against friction is

  • 200 J

  • 100 J

  • Zero

  • 1000 J

Solution

B.

100 J

Net work done in sliding a body up to a height h on inclined plane
    = Work done against gravitational force + Work done against frictional force
rightwards double arrow space space space space space straight W space equals space straight W subscript straight g plus straight W subscript straight f space space space space space space space space space space space... left parenthesis straight i right parenthesis
But space space space space space straight W space equals space 300 space straight J
space space space space space space straight W subscript straight g space equals space mgh space equals space 2 space cross times 10 cross times 10 space equals space 200 space straight J
putting space in space Eq. space left parenthesis straight i right parenthesis comma space we space get
space space space space space space space space space 300 space equals space 200 plus straight W subscript straight f
rightwards double arrow space space space space space space space space space space space space space straight W subscript straight f space equals space 300 space minus space 200 space equals space 100 space straight J
 

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