Laws of Motion

Question
CBSEENPH11020772

A block B is pushed momentarily along a horizontal surface with an initial velocity v. if μ is the coefficient of sliding friction between B  and the surface, block B will come to rest after a time:

  • v/gμ

  • gμ/v

  • g/v

  • v/g

Solution

A.

v/gμ

Block B will come to rest, if force applied to it will vanish due to frictional force acting between block B and surface, ie,
force applied = frictional force
i.e., μmg = ma
μmg = m (v/t)
t =v/μg

Question
CBSEENPH11020795

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking =10 m/s2, work done against friction is

  • 200 J

  • 100 J

  • Zero

  • 1000 J

Solution

B.

100 J

Net work done in sliding a body up to a height h on inclined plane
    = Work done against gravitational force + Work done against frictional force
rightwards double arrow space space space space space straight W space equals space straight W subscript straight g plus straight W subscript straight f space space space space space space space space space space space... left parenthesis straight i right parenthesis
But space space space space space straight W space equals space 300 space straight J
space space space space space space straight W subscript straight g space equals space mgh space equals space 2 space cross times 10 cross times 10 space equals space 200 space straight J
putting space in space Eq. space left parenthesis straight i right parenthesis comma space we space get
space space space space space space space space space 300 space equals space 200 plus straight W subscript straight f
rightwards double arrow space space space space space space space space space space space space space straight W subscript straight f space equals space 300 space minus space 200 space equals space 100 space straight J
 

Question
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
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