Motion in Straight Line

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Question
CBSEENPH11020683

A ball is dropped from  a high-rise platform at t=0 starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t= 18s. What is the value of v? (take g= 10 ms-2)

  • 74 ms-2

  • 55 ms-1

  • 40 ms-1

  • 60 ms-1

Solution

A.

74 ms-2

For space first space ball space comma space straight u space equals space 0
therefore comma
straight s subscript 1 space equals space 1 half space gt subscript 1 superscript 2 space equals space 1 half space straight x space straight g space left parenthesis 18 right parenthesis squared
For space second space ball comma space initial space velocity space equals space straight v
therefore comma space
straight s subscript 2 space equals space vt subscript 2 space plus space 1 half space gt squared
straight t subscript 2 space equals space 18 minus 6 space equals space 12 space straight s
straight s subscript 2 space equals space straight v space straight x space 12 space plus space 1 half straight g space left parenthesis 12 right parenthesis squared
Here comma space space straight s subscript 1 space equals space straight s subscript 2
1 half straight g left parenthesis 18 right parenthesis squared space equals space 12 space straight v space plus space 1 half space left parenthesis straight g right parenthesis space left parenthesis 12 right parenthesis squared
straight v equals space 74 space ms to the power of negative 1 end exponent

Sponsor Area

Question
CBSEENPH11020792

A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the relation straight s equals 1 third straight t squared, where t is in s. Work done by the force in 2 s is:

  • 5 over 19 straight J
  • 3 over 8 straight J
  • 8 over 3 straight J
  • 19 over 5 straight J

Solution

C.

8 over 3 straight J

If a constant force is applied on the object causing a displacement in it, then it is said that work has been done on the body to displace it.
  Work done by the force  = Force x Displacement
or          W = F x s                            ...(i)
But from Newton's 2nd law, we have
  Force  =  Mass x Acceleration
i.e.,       F = ma                                ...(ii)
Hence, from Eqs. (i) and (ii), we get
straight W space equals space mas space equals space straight m open parentheses fraction numerator straight d squared straight s over denominator dt squared end fraction close parentheses straight s space space space space space space... left parenthesis iii right parenthesis space space space open parentheses because space straight a space equals space fraction numerator straight d squared straight s over denominator dt squared end fraction close parentheses
Now comma space we space have comma space space straight s space equals space 1 third straight t squared
therefore space space space space space space space space space fraction numerator straight d squared straight s over denominator dt squared end fraction space equals space straight d over dt open square brackets straight d over dt open parentheses 1 third straight t squared close parentheses close square brackets
space space space space space space space space space space space space space space space space space space space equals straight d over dt cross times open parentheses 2 over 3 straight t close parentheses
space space space space space space space space space space space space space space space space space space equals 2 over 3 dt over dt space space space space space space
space space space space space space space space space space space space space space space space space space equals 2 over 3 space space space space space space space
space space space space space space space space space space space space space space space
Hence, Eq (ii) becomes
                straight W space equals 2 over 3 ms space equals space 2 over 3 straight m cross times 1 third straight t squared
                   equals 2 over 9 mt squared
We have given
                   straight m equals 3 space kg comma space straight t space equals space 2 straight s
therefore space space space space straight w space equals space 2 over 9 cross times 3 cross times left parenthesis 2 right parenthesis squared space equals space 8 over 3 straight J
 

Question
CBSEENPH11020673

A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is μ = 0.5. The distance that the box - will move relative to belt before coming to rest on it taking 

g= 10 ms-2 is

  • 1.2 m

  • 0.6 m

  • zero

  • 0.4 m

Solution

D.

0.4 m


Force, 
F = μg
Retardation of the block on the belt
straight a space equals space straight F over straight m space equals space fraction numerator straight mu space mg over denominator straight m end fraction space equals space μg

From comma space straight v squared space equals space straight u squared space plus 2 space as
0 space equals space left parenthesis 2 right parenthesis squared space minus space 2 space left parenthesis μg right parenthesis straight s

straight s space equals space fraction numerator 4 over denominator 2 space straight x space 0.5 space straight x space 10 end fraction space equals space 0.4 space straight m

Question
CBSEENPH11020672

A particle covers half of its total distance with speed v1 and the rest half distance with speed v2. Its average speed during the complete journey is

  • v1v/ v1 + v2

  • 2v1v2/v1 + v2

  • 2v12v22 / v12 + v22

  • v1 + v2 / 2

Solution

B.

2v1v2/v1 + v2

Velocity v = s / t
s = vt
The average speed of particle
straight v subscript av space equals space fraction numerator straight s space plus space straight s over denominator begin display style straight s over straight v subscript 1 end style plus begin display style straight s over straight v subscript 2 end style end fraction

straight v subscript av space equals space fraction numerator 2 straight v subscript 1 straight v subscript 2 over denominator straight v subscript 1 space plus straight v subscript 2 end fraction

Question
CBSEENPH11020564

A particle is moving such that its position coordinates (x,y) are (2m, 3m) at time t = 0, (6m, 7m) at time t = 2 sec and (13 m, 14m) at time t= 5 sec. Average velocity vector (vav) from t = 0 to t = 5 s is,

  • 1 fifth left parenthesis 13 space i with hat on top space plus space 14 space j with hat on top right parenthesis
  • 7 over 3 open parentheses i with rightwards arrow on top space plus space j with hat on top close parentheses
  • 2 space left parenthesis straight i with hat on top space plus space straight j with hat on top right parenthesis
  • 11 over 5 open parentheses i with hat on top plus j with hat on top close parentheses

Solution

D.

11 over 5 open parentheses i with hat on top plus j with hat on top close parentheses

Average velocity vector,

straight v subscript av space equals space fraction numerator Net space displacement over denominator Time space taken end fraction

space space space space space space equals space fraction numerator left parenthesis 13 minus 2 right parenthesis straight i with hat on top space plus space left parenthesis 14 minus 3 right parenthesis straight j with hat on top over denominator 5 end fraction

space space space space space space equals space fraction numerator 11 space straight i with hat on top space plus space 11 space straight j with hat on top over denominator 5 end fraction

space space space space space space space equals space 11 over 5 open parentheses straight i with hat on top space plus space straight j with hat on top close parentheses