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

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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
WiredFaculty
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
                   WiredFaculty
 

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

WiredFaculty
Force, 
F = μg
Retardation of the block on the belt
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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
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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,

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15