Laws of Motion

Question 509
CBSEENPH11020468

A machine gun fires a bullet of mass 40 g with a velocity 1200 ms−1. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most?

  • one 

  • Four

  • Two

  • Three

Solution

D.

Three

The force exerted by machine gun on man's hand in firing a bullet = change in momentum per second on a bullet or rate of change of momentum
space equals open parentheses 40 over 1000 close parentheses space straight x space 1200 space equals space 48 space straight N
The force exerted by man on machine gun = 144 N Hence, number of bullets fired =144/48 = 3

Question 510
CBSEENPH11020470

A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (take g = 10 m/s2 )
  • 2.0

  • 4.0

  • 1.6 

  • 2.4

Solution

A.

2.0

Let the mass of block be m.
Frictional force in rest position
F = mg sin 30

10 space equals space straight m space straight x space 10 space straight x space 1 half
space straight m space equals space fraction numerator 2 space straight x space 10 over denominator 10 end fraction space equals space 2 space kg

Question 511
CBSEENPH11020501

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :

  • (0,1)

  • (.89,.28)

  • (.28,.89)

  • (0,0)

Solution

B.

(.89,.28)

For collision of a neutron with deuterium:

Applying conservation of momentum:

mv + 0 = mv1 + 2mv2 .....(i)
v2 -v1 = v ...... (ii)

Therefore, Collision is elastic, e = 1

From equ (i) and equ (ii) v1 = -v/3

Pd = 12mv2 -12mv1212mv2 = 89 = 0.89

Now, for the collision of neutron with carbon nucleus

Applying conservation of momentum

mv + 0 = mv1 + 12mv2 ....; (iii)

v = v2-v1  ....(iv)

v1 = -1113 vPc = 12mv2 - 12m1113v212mv2 = 48169 0.28

Question 512
CBSEENPH11020513

A car is negotiating a curved road of radius R. The road is banked at angle straight theta. The coefficient of friction between the tyres of the car and the road is straight mu subscript straight s. The maximum safe velocity on this road is,

  • square root of gR open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tan space straight theta end fraction close parentheses end root
  • square root of straight g over straight R open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tanθ end fraction close parentheses end root
  • square root of straight g over straight R squared open parentheses fraction numerator mu subscript s plus tan theta over denominator 1 minus mu subscript s tan theta end fraction close parentheses end root
  • square root of g R squared open parentheses fraction numerator straight mu subscript straight s plus tanθ over denominator 1 minus straight mu subscript straight s tanθ end fraction close parentheses end root

Solution

A.

square root of gR open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tan space straight theta end fraction close parentheses end root

A car is negotiating a curved road of radius R. The road is banked at angle straight theta and the coefficient of friction between the tyres of car and the road is straight mu subscript straight s.
The given situation is illustrated as:

In the case of vertical equilibrium,
N cos straight theta = mg + f1 sin straight theta
rightwards double arrowmg = N cos straight theta space minus space straight f subscript 1 space sin space straight theta    ... (i)
In the case of horizontal equilibrium,
straight N space sin space straight theta space plus space straight f subscript 1 space cosθ space equals space mv squared over straight R  ... (ii)
Dividing Eqns. (i) and (ii), we get
straight v squared over Rg equals fraction numerator sin space theta space plus space mu subscript s space cos theta over denominator cos space theta space minus space mu subscript s space sin space theta end fraction space open square brackets f subscript 1 proportional to mu subscript s close square brackets

rightwards double arrow v space equals square root of R g open parentheses fraction numerator tan space theta space plus mu subscript s over denominator cos space theta space minus space mu subscript s space sin space theta space end fraction close parentheses end root
rightwards double arrow space v space equals space square root of R g open parentheses fraction numerator tan space theta space plus space mu subscript s over denominator 1 minus mu subscript s space tan space theta end fraction close parentheses end root

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