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Laws of Motion Test - 23

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Laws of Motion Test - 23
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  • Question 1
    1 / -0
    kg ms$$^{-1}$$ is the SI unit of
    Solution
    Impulse $$(I)$$ imparted to the body is equal to the change in linear momentum $$(\Delta p)$$ of the body.
    Impulse  $$I = \Delta p$$
    Thus momentum has the same units as that of impulse.
    Since momentum has SI unit as  $$kg \ m/s$$, thus impulse also has SI unit as $$kg \ m/s$$.
  • Question 2
    1 / -0
    A body of mass $$5  kg$$ undergoes a change in speed form $$30 m/s$$ to $$40  {m}/{s}$$. Its momentum would increase by
    Solution
    change in momentum = 5 (40 - 30) = $$50kgm/s$$
  • Question 3
    1 / -0
    Newton used the phrase 'quantity of motion' for
    Solution
    quantity of motion in a body, the relative amount of its motion, as measured by its momentum, varying as the product of mass and velocity.which is define as momentum.
  • Question 4
    1 / -0
    Momentum has the same units as that of 
    Solution
    Impulse applied to an object produces an equivalent vector change in its linear momentum, also in the same direction. The SI unit of impulse is the newton second $$(N⋅s)$$, and the dimensionally equivalent unit of momentum is the kilogram meter per second $$(kg⋅m/s).$$
  • Question 5
    1 / -0
    A number of discs, each of momentum $$M  kg  {m}/{s}$$ are striking a wall at the rate of $$n$$ discs per minute. The force associated with these discs, in newtons, would be
    Solution
    The force is $$F=\dfrac{\Delta P}{dt}$$
    $$momentum=M kg.m/s$$
    change in time=rate of dics striking wall 
    $$\Delta t=\dfrac{1}{n/minute}=\dfrac{60}{n}$$ 
    $$F=\dfrac{\Delta P}{dt}=\dfrac{M}{60/n}=\dfrac{Mn}{60}N$$
  • Question 6
    1 / -0
    A body of mass $$5  kg$$ undergoes a change in speed from $$20$$ to $$0.20  {m}/{s}$$. The momentum of the body would
    Solution
    $$momentum=mass*\delta v=5kg*(20-0.20)=99kgm/s$$
    Hence it decreases by $$99kgm/s$$
  • Question 7
    1 / -0
    A body of mass $$0.1 kg$$ is moving with a velocity of $$15  {m}/{s}$$. The momentum associated with the ball will be
    Solution
    P =mv = 0.1 x 15 $$kgm/s$$ = 1.5$$kgm/s$$
  • Question 8
    1 / -0
    The combined effect of mass and velocity is taken into account by a physical quantity called 
    Solution
    P = mv
    i.e it takes in account the factors of both m and v .
  • Question 9
    1 / -0
    A long-jumper runs before jumping because
    Solution
    to gain momentum so that the range that he jumps is long
  • Question 10
    1 / -0
    The force acting on an object are balanced, then the object:
    Solution
    Balanced forces always result in a zero external force acting on an object  i.e  $$F_{ext}  =0$$
    $$\therefore$$   Acceleration       $$a  = \dfrac{F_{ext}}{m}  =  0$$
    Thus the object must not be accelerating or it must be moving with constant velocity or remain at rest (if it was at rest initially). Hence option C is correct.
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