Laws of Motion

Question 545
CBSEENPH11020796

A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 s, the average force acting on the wall is

  • 48 N

  • 24 N

  • 12 N

  • 96 N

Solution

B.

24 N

The vector OA with rightwards arrow on top represents the momentum of the object before the collision, and the vector OB with rightwards arrow on top that after the collision. The vector AB with rightwards arrow on top represents the change in momentum of the object increment straight P with rightwards arrow on top.


As the magnitudes of  OA with rightwards arrow on top space and space OB with rightwards arrow on top are equal, the components of OA with rightwards arrow on top space and space OB with rightwards arrow on top along the wall are equal and in the same direction, while those perpendicular to the wall are equal and opposite. Thus, the change in momentum is due only to the change in direction of the perpendicular components.
Hence,  increment straight p space equals space OB space sin space 30 degree space minus space left parenthesis negative OA space sin space 30 degree right parenthesis
                    equals mv space sin space 30 degree space minus space left parenthesis negative mv space sin space 30 degree right parenthesis
equals 2 space mv space sin space 30 degree
Its time rate will appear in the form of average force acting on the wall.
 therefore space space space space space space space space space straight F space cross times space straight t space equals space 2 mv space sin space 30 degree
or space space space space space space space space space space space space space space straight F space equals space fraction numerator 2 mv space sin space 30 degree over denominator straight t end fraction
   Given comma space straight m space equals space 0.5 space kg comma space space straight v space equals space 12 space straight m divided by straight s comma space space space straight t equals space 0.25 space straight s
space space space space space space space space space space straight theta space equals space 30 degree
Hence comma space space space straight F space equals space fraction numerator 2 cross times 0.5 cross times 12 space sin space 30 degree over denominator 0.25 end fraction space equals space 24 space straight N

Question 546
CBSEENPH11020797

The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc is

  • MR squared
  • 2 over 5 MR squared
  • 3 over 2 MR squared
  • 1 half MR squared

Solution

C.

3 over 2 MR squared

The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is
                     straight I subscript CM space equals space 1 half MR squared
where M is the mass of disc and R its radius. According to theorem of parallel axis, MI of circular disc about an axis touching the disc at it diameter and normal to the disc is
                   straight I equals space straight I subscript CM space plus space MR squared
                     equals 1 half MR squared plus space MR squared
equals 3 over 2 MR squared

Question 547
CBSEENPH11020806

A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: (Moment of inertia of rod about A is fraction numerator straight m l squared over denominator 3 end fraction)

  • fraction numerator 3 straight g over denominator 2 straight l end fraction
  • fraction numerator 2 straight l over denominator 3 straight g end fraction
  • fraction numerator 3 straight g over denominator 2 straight l squared end fraction
  • mg straight l over 2

Solution

A.

fraction numerator 3 straight g over denominator 2 straight l end fraction

The moment of inertia of the uniform rod about an axis through the end and perpendicular to length is
           straight I space equals space fraction numerator straight m l squared over denominator 3 end fraction
where m is mass of rod and l its length.
Torque left parenthesis straight t space equals space Iα right parenthesis acting on centre of gravity of rod is given by
                  straight t equals mg l over 2
or               Iα space equals space mg l over 2
or space space fraction numerator straight m l squared over denominator 3 end fraction straight alpha space equals space mg l over 2
therefore space space space space straight alpha space equals fraction numerator 3 straight g over denominator 2 l end fraction

 

Question 548
CBSEENPH11020820

Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]

  • 350 Hz

  • 311 Hz

  • 411 Hz

  • 448 Hz

Solution

D.

448 Hz

straight f subscript straight A space equals space straight f open square brackets fraction numerator straight v plus straight v subscript 0 over denominator straight v minus straight v subscript straight s end fraction close square brackets
space equals space 400 space open square brackets fraction numerator 340 space plus 16.5 over denominator 340 minus 22 end fraction close square brackets
straight f subscript straight A space equals space 448 space Hz

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