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Forces and Laws of Motion Test - 52

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Forces and Laws of Motion Test - 52
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  • Question 1
    1 / -0
    In the graph below is given the variation of force with time. Find out the net change in momentum of the object.

    Solution

  • Question 2
    1 / -0
    From a stationary tank of mass $$125000$$ pound, a small shell of mass $$25$$ pound is fired with a muzzle velocity of $$1000\ ft/sec$$. The tank recoils with a velocity of ..........
    Solution
    Given:
    Mass of the tank $$M=125000\ pound$$
    Mass of the shell $$m=25\ pound$$
    Velocity of the shell $$v=100\ ft/sec$$
    Recoil velocity of the tank $$V=?$$
    Since There is no external force acting, we can apply conservation of momentum.
    According to the conservation of momentum,
    $$\text{Initial momentum of (tank+shell)} =\text{Final momentum of (tank + shell)}$$ 
    $$(M+m)\times 0 =M\times V + m\times v$$
    $$\Rightarrow 0 = 125000\times V + 25\times 1000$$
    $$\Rightarrow V=-\dfrac{25000}{125000}=-\dfrac{1}{5}=-0.2\ ft/sec$$
    The -ve sign indicates that the tank will recoin in the opposite direction to the direction of motion of the shell.

  • Question 3
    1 / -0
    When a body is stationary 
    Solution

    Consider a stationary block on a table. Gravity force (weight $$W=mg$$) acts downwards and normal reaction force from table acts upwards.

    Hence, the combination of forces acting on it balance each other.                  

  • Question 4
    1 / -0
    When two bodies A and B interact with each other, A exerts a force of $$10N$$ on B towards east. What is the force exerted by B on A?
    Solution
    From Newton's third law of motion, "Every action have equal and opposite reaction".
    Here, Body A is exerting 10 N force on B towards east. (Action force)
    hence Body B will apply same amount of reaction force ie 10 N on A in direction opposite of east ie in west.
    Option A
  • Question 5
    1 / -0
    In accordance with Newton's third law of motion 
    Solution
    A force is a push or a pull that acts upon an object as a result of its interaction with another object. These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction.
  • Question 6
    1 / -0
     A person in a car tends to fall back when it suddenly starts. It is due to
    Solution
    When the bus suddenly starts moving in forward direction. This happens due to. When a bus suddenly starts, the standing passengers fall backward in the bus.
    It is due to inertia of rest.
    Hence,
    option $$A$$ is correct answer.
  • Question 7
    1 / -0
    Newton's $$2nd$$ law explain us
    Solution
     Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
    Hence,
    option $$A$$ is correct answer.
  • Question 8
    1 / -0
    Rahul is traveling in a bus on a straight road facing in the direction of the motion of the bus. If the bus starts accelerating in the same direction then Rahul will fall 
    Solution
    The face is in the direction of motion so the force due to the acceleration of Bus will be opposite to motion that is opposite to face that is backward so he will fall backward.
    The instant acceleration of the bus will be applied only to the foot of Rahul, while his upper body remains in the previous state, so Rahul will fall backward.
  • Question 9
    1 / -0
    A boy who normally weigh $$200 \ N$$ on a spring scale suddenly jump upwards. The scale reading jumps to $$400 \ N$$. What is boy's maximum acceleration upwards? $$(g=9.8m/s^2)$$
    Solution
    Given,
    Weight $$W= 200 \ N$$
    $$g=9.8m/s^2$$

    The Normal weight of the boy,
    $$W=mg$$

    $$m=\dfrac{W}{g}=\dfrac{200}{9.8}$$
    $$m=20.40kg$$

    When a boy suddenly jump upward with acceleration $$a$$, then reading is $$400N$$
    This reading is nothing but the normal force on the boy in upward direction. 
    So two forces act on the boy, one is $$mg$$ downwards and the other is $$N=400 \ N$$ upwards. The resultant of these forces gives an acceleration $$a$$ in the upward direction.
    By applying Newton's second law,
    $$N-mg=ma$$ (taking up as positive)
    $$400-20.40\times 9.8=20.40a$$
    $$20.40a=400-199.92$$
    $$20.40a=200.08$$

    $$a=\dfrac{200.8}{20.40}$$

    $$a=9.8m/s^2$$

    The correct option is C.

  • Question 10
    1 / -0
    A car is moving with $$15\ ms^{-1}$$ whose mass is $$1500\ kg$$. Find the momentum of the car.
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