A man of 50 kg mass standing in a gravity free space at height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 ms-1. When the stone reaches the floor, the distance of the man above the floor will be

  • 9.9 m

  • 10.1 

  • 10 m

  • 10 m


B.

10.1 

m r = constant



The distance of the man above the floor (total height) = 10+0.1 = 10.1

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A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance  and the time are e1 and e2 respectively, the percentage error in the estimation of g is

  • e1 - e2

  • e1 + 2e2

  • e1+ e2

  • e1+ e2


B.

e1 + 2e2

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An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are 1 kg first part moving with a velocity of 12 ms-1 and 2 kg second part moving with a velocity of 8 ms-1.If the third part flies off with a velocity of 4 ms-,its mass would be

  • 5 kg 

  • 7 kg

  • 17 kg

  • 17 kg


A.

5 kg 

apply the law of conservation of linear momentum. 

momentum of first part = 1 x 12 = 12 kg ms-1
Momentum of the second part  = 2 x 8 = 16 kg ms-1 '

Resultant monmentum

= [(12)2 +(16)2]1/2 = 20 kg ms-1

The third part should also have the same momentum.


Let the mass of third part be M, then 

4 x M = 20

M = 5 kg
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A body under the action of a force,straight F with rightwards arrow on top space equals space 6 space bold i with bold hat on top space minus space 8 space bold j with bold hat on top space plus 10 space bold k with hat on top acquires an acceleration of 1 ms-2. The mass of this body must be

  • 2 square root of 10 space kg
  • 20 kg

  • 10 kg

  • 10 kg


D.

10 kg

According to Newton's second law of motion, force = mass x acceleration.
Here, 

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The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is 

  • 30 ms-2 downward

  • 4 ms-2  upwards

  •  4 ms-2  downwards

  •  4 ms-2  downwards


B.

4 ms-2  upwards

Apparent weight > actual weight, then the lift is accelerating upward.

Lift is accelerating upward at the rate of a 
Hence, equation of motion is written as
R - mg = ma

28000-20000 = 2000a

a = 8000/2000 = 4 ms-2 upwards.

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