Motion in A Plane

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Question
CBSEENPH11020783

bold A with bold rightwards arrow on top bold space and bold space bold B with bold rightwards arrow on top are two vectors and θ is the angle between them if vertical line bold A with bold rightwards arrow on top bold space straight x bold space bold B with bold rightwards arrow on top bold vertical line bold space equals bold space square root of bold 3 bold space bold left parenthesis bold A with bold rightwards arrow on top bold. bold B with bold rightwards arrow on top bold right parenthesis the value of θ is:
  • 60o

  • 45o

  • 30o

  • 90o

Solution

A.

60o

bold A with bold rightwards arrow on top bold space bold x bold space bold B with bold rightwards arrow on top bold space equals space AB space sin space space and space stack bold A bold. with bold rightwards arrow on top bold B with bold rightwards arrow on top bold space equals space AB space cos space straight theta.
Given comma space vertical line bold A with bold rightwards arrow on top bold space bold x bold space bold B with bold rightwards arrow on top vertical line space equals space square root of 3 open parentheses stack bold A bold. with bold rightwards arrow on top bold B with bold rightwards arrow on top bold space close parentheses
rightwards double arrow space AB space sin space space straight theta space equals square root of 3 space AB space cos space space straight theta
tan space equals space square root of 3
space straight theta space equals space 60 to the power of 0

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Question
CBSEENPH11020632

A  particle of mass m is released from rest and follows a parabolic path is shown. Assuming that the displacement of the mass from the origin is small, which graph correctly depicts the position of the particle as a function of time?

Solution

D.

Potential V (x) versus x is parabolic, SHM starts from an extreme position and x versus t should be a cosin curve.

Question
CBSEENPH11020568

A balloon with mass m is descending down with an acceleration a (where a <g). How much mass should be removed from it so that it starts moving up with an acceleration a?

  • fraction numerator 2 ma over denominator straight g plus straight a end fraction
  • fraction numerator 2 ma over denominator straight g minus straight a end fraction
  • fraction numerator ma over denominator straight g plus straight a end fraction
  • fraction numerator ma over denominator straight g minus straight a end fraction

Solution

A.

fraction numerator 2 ma over denominator straight g plus straight a end fraction

When the balloon is descending down with acceleration a,

So, mg - B = mx A             ... (i)
where B is the buoyant force.
We assume here that while removing same mass, the volume of balloon and hence buoyant force will not change.
Let, us assume the new mass of the balloon is m'.
So, mass removed is (m-m')
Therefore,
B-m'g = m'x a         ... (ii)
On solving equations (i) and (ii), we have
mg - B = m x a
B - m'g = m' x a
mg - m'g = ma +  m'a
(mg - ma) = m' (g+a) = m (g-a) = m' (g+a)
That is,
straight m apostrophe space equals space fraction numerator straight m space left parenthesis straight g minus straight a right parenthesis over denominator straight g plus straight a end fraction
That is, mass removed is m-m'
equals space straight m space open square brackets fraction numerator 1 minus left parenthesis straight g minus straight a right parenthesis over denominator left parenthesis straight g plus straight a right parenthesis end fraction close square brackets
equals space straight m open square brackets fraction numerator left parenthesis straight g plus straight a right parenthesis minus left parenthesis straight g minus straight a right parenthesis over denominator left parenthesis straight g plus straight a right parenthesis end fraction close square brackets
equals space straight m open square brackets fraction numerator straight g plus straight a minus straight g plus straight a over denominator straight g plus straight a end fraction close square brackets
equals space increment straight m space equals space fraction numerator 2 ma over denominator straight g plus straight a end fraction

Question
CBSEENPH11020637

A body is moving with velocity 30 m/s towards east. After 10 s its velocity becomes 40 m/s towards the north. The average acceleration of the body is 

  • 7 m/s2

  • square root of 7 space m divided by s squared
  • 5 m/s2

  • 1 m/s2

Solution

C.

5 m/s2

Average space acceleration space equals space fraction numerator Change space in space velocity over denominator Total space time end fraction

straight a space equals space fraction numerator vertical line straight v subscript straight f space minus space straight v subscript straight i vertical line over denominator increment straight t end fraction

equals space fraction numerator square root of 30 squared plus 40 squared end root over denominator 10 end fraction space equals space fraction numerator square root of 900 plus 1600 end root over denominator 10 end fraction
space equals space 5 space ms to the power of negative 2 end exponent

Question
CBSEENPH11020514

A body of mass 1 kg begins to move under the action of a time dependent force F = left parenthesis 2 straight t space straight i with hat on top space plus space 3 straight t squared space straight j with hat on top right parenthesis N, where straight i with hat on top space a n d space j with hat on top are units vectors along X and Y axis. What power will be developed by the force at the time (t)?

  • (2t2 + 4t4) W

  • (2t3 + 3 t4) W

  • (2t3 + 3t5) W

  • (2t + 3t3)W

Solution

C.

(2t3 + 3t5) W

A body of mass 1 kg begins to move under the action of time dependent force,
F = (2t straight i with hat on top+3t2 straight j with hat on top) N,
where stack straight i space with hat on top a n d space j with hat on top are unit vectors along X and Y axis.
F = ma
rightwards double arrow space straight a space equals space straight F over straight m
rightwards double arrow space straight a space equals space fraction numerator left parenthesis 2 straight t space straight i with hat on top space plus space 3 straight t squared space straight j with hat on top right parenthesis over denominator 1 end fraction space space left square bracket space straight m space equals 1 space kg right square bracket
rightwards double arrow space straight a space equals space left parenthesis 2 straight t space straight i with hat on top space plus space 3 straight t squared space straight j with hat on top right parenthesis space straight m divided by straight s squared
Acceleration, a = dv over dt
rightwards double arrow space dv space equals space straight a. space dt space space space space space space... space left parenthesis straight i right parenthesis
Integrating both sides, we get
integral d v space equals space integral a. space d t
space space space space space space space space space equals space integral left parenthesis 2 t space i with hat on top space plus space 3 t squared space j with hat on top right parenthesis space d t
v space equals space t squared space space i with hat on top space plus straight t cubed space straight j with hat on top
Power developed by the force at the time t will be given as,
P = F.v = (2t straight i with hat on top + 3t2 straight j with hat on top).(straight t squared space straight i with hat on top space plus space straight t cubed space straight j with hat on top)
   = (2t. t2 + 3t2.t3)
P = (2t3 + 3t5) W

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