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CBSE-XI-Physics

03: Rest and Motion: Kinematics

with Solutions - page 2

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  • Qstn #11
    In figure shows the x coordinate of a particle as a function of time. Find the sings of vx and ax at t = t1, t = t2 and t = t3.
    Figure
    Ans : Slope of the x-t graph gives the velocity and change in the slope gives the acceleration.
    At t = t1,
    Slope = Positive ⇒ Velocity = Positive
    Slope is increasing ⇒ Acceleration = Positive
    At t = t2,
    Slope = Zero ⇒ Velocity = Zero
    Slope is increasing ⇒ Acceleration = Negative
    At t = t3,
    Slope = Negative ⇒ Velocity = Negative
    Slope is increasing ⇒ Acceleration = Positive
  • Qstn #12
    A player hits a baseball at some angle. The ball goes high up in space. The player runs and catches the ball before it hits the ground. Which of the two (the player or the ball) has greater displacement?
    Ans : The displacement of the ball and the player is the same, as the initial and final points are the same.
  • Qstn #13
    The increase in the speed of a car is proportional to the additional petrol put into the engine. Is it possible to accelerate a car without putting more petrol or less petrol into the engine?
    Ans : Yes, it is possible to accelerate a car without putting more petrol or less petrol in the engine. This can be done by driving the car on a circular or curved track at a uniform speed.
  • Qstn #14
    Rain is falling vertically. A man running on the road keeps his umbrella tilted but a man standing on the street keeps his umbrella vertical to protect himself from the rain. But both of them keep their umbrella vertical to avoid the vertical sun-rays. Explain.
    Ans : The speed of rain is less, so as we run, the direction of the relative velocity of the rain changes. But as the speed of light is very high, there is a measurable change in the relative velocity of light w.r.t. a person due to the relative motion between the sunrays and the person.
  • #
    Section : B
    MCQS
  • Qstn #1
    A motor car is going due north at a speed of 50 km/h. It makes a 90° left turn without changing the speed. The change in the velocity of the car is about
    -(a) 50 km/h towards west
    (b) 70 km/h towards south-west
    (c) 70 km/h towards north-west
    (d) zero.
    digAnsr:   b
    Ans : (b) 70 km/h towards south-west
    Final velocity, `` \stackrel{\to }{{V}_{f}}=-50\stackrel{‸}{i}\,\mathrm{\,km \,}/\,\mathrm{\,h \,}``
    Initial velocity, `` \stackrel{\to }{{V}_{i}}=50\stackrel{‸}{j}\,\mathrm{\,km \,}/\,\mathrm{\,h \,}``
    Change in velocity, `` ∆\stackrel{\to }{V}=\stackrel{\to }{{V}_{f}}-\stackrel{\to }{{V}_{i}}``
    `` \left|∆V\right|=\sqrt{{50}^{2}+{50}^{2}+2\times 50\times 50\,\mathrm{\,cos \,}\left(90°\right)}=70\,\mathrm{\,km \,}/\,\mathrm{\,h \,}``
    It is towards southwest, as shown in the figure.
  • Qstn #2
    In figure shows the displacement-time graph of a particle moving on the X-axis.
    Figure
    -(a) the particle is continuously going in positive x direction
    (b) the particle is at rest
    (c) the velocity increases up to a time t0, and them become constant
    (d) the particle moves at a constant velocity up to a time t0, and then stops.
    digAnsr:   d
    Ans : (d) The particle moves at a constant velocity up to a time t0 and then stops.
    The slope of the x-t graph gives the velocity. In the graph, the slope is constant from t = 0 to t = t0, so the velocity is constant. After t = t0, the displacement is zero; i.e., the particle stops.
  • Qstn #3
    A particle has a velocity u towards east at t = 0. Its acceleration is towards west and is constant. Let xA and xB be the magnitude of displacements in the first 10 seconds and the next 10 seconds
    -(a) xA < xB
    (b) xA = xB
    (c) xA > xB
    (d) the information is insufficient to decide the relation ofxA with xB.
    digAnsr:   d
    Ans : (d) The information is insufficient to decide the relation of xA with xB.
    As velocity and acceleration are in opposite directions, velocity will become zero after some time (t) and the particle will return.
    `` \therefore 0=u-at``
    `` \Rightarrow t=\frac{u}{a}``
    Because the value of acceleration is not given, we cannot say that the particle will return after/before 10 seconds.
  • Qstn #4
    A person travelling on a straight line moves with a uniform velocity v1 for some time and with uniform velocity v2 for the next equal time. The average velocity v is given by
    -(a)
    v=v1+v22
    (b)
    v=v1v2
    (c)
    2v=1v1+1v2
    (d)
    1v=1v1+1v2
    digAnsr:   a
    Ans : (a) `` v=\frac{{v}_{1}+{v}_{2}}{2}``
    Velocity is uniform in both cases; that is, acceleration is zero.
    We have:
    `` {d}_{1}={v}_{1}t`` and `` {d}_{2}={v}_{2}t``
    Total displacement, `` d={d}_{1}+{d}_{2}``
    Total time, `` t=t+t=2t``
    `` \therefore `` Average velocity, `` v=\frac{{d}_{1}+{d}_{2}}{2t}=\frac{{v}_{1}+{v}_{2}}{2}``
  • Qstn #5
    A person travelling on a straight line moves with a uniform velocity v1 for a distance x and with a uniform velocityv2 for the next equal distance. The average velocity v is given by
    -(a)
    v=v1+v22
    (b)
    v=v1v2
    (c)
    2v=1v1+1v2
    (d)
    1v=1v1+1v2
    digAnsr:   c
    Ans : (c) `` \frac{2}{v}=\frac{1}{{v}_{1}}+\frac{1}{{v}_{2}}``
    Velocity is uniform in both cases; that is, acceleration is zero.
    `` x={v}_{1}{t}_{1}\Rightarrow {t}_{1}=\frac{x}{{v}_{1}}``
    `` x={v}_{2}{t}_{2}\Rightarrow {t}_{2}=\frac{x}{{v}_{2}}``
    Total displacement, `` x\text{ ' }=2x``
    Total time, `` t={t}_{1}+{t}_{2}``
    `` \therefore `` Average velocity, `` v=\frac{x\text{ ' }}{t}=\frac{2{v}_{1}{v}_{2}}{{v}_{1}+{v}_{2}}``
    `` \Rightarrow \frac{2}{v}=\frac{1}{{v}_{1}}+\frac{1}{{v}_{2}}``
  • Qstn #6
    A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is
    -(a) a upward
    (b) (g - a) upward
    (c) (g - a) downward
    (d) g downward
    digAnsr:   d
    Ans : (d) g downward
    Gravity is the only force acting on the stone when it is released. And, we know that gravity is always in the downward direction.
  • Qstn #7
    A person standing near the edge of the top of a building throws two balls A and B. the ball A is thrown vertically upward and B is thrown vertically downward with the same speed. The ball A hits the ground with a speed vA and the ball B this the ground with a speed vB. We have
    -(a) vA > vB
    (b) vA < vB
    (c) vA = vB
    (d) the relation betweenvA andvB depends on height of the building above the ground.
    digAnsr:   c
    Ans : (c) vA = vB
    Total energy of any particle = `` \frac{1}{2}m{v}^{2}+mgh``
    Both the particles were at the same height and thrown with equal initial velocities, so their initial total energies are equal. By the law of conservation of energy, their final energies are equal.
    At the ground, they are at the same height. So, their P.E. are also equal; this implies that their K.E. should also be equal. In other words, their final velocities are equal.
  • Qstn #8
    In a projectile motion the velocity
    -(a) is always perpendicular to the acceleration
    (b) is never perpendicular to the acceleration
    (c) is perpendicular to the acceleration for one instant only
    (d) is perpendicular to the acceleration for two instants.
    digAnsr:   c
    Ans : (c) is perpendicular to the acceleration for one instant only
    In projectile motion, velocity is perpendicular to acceleration only at the highest point. Here, velocity is along the horizontal direction and acceleration is along the vertically downward direction.
  • Qstn #9
    Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit he ground first?
    -(a) the faster one
    (b) the slower one
    (c) both will reach simultaneously
    (d) depends on the masses.
    digAnsr:   c
    Ans : (c) Both will reach simultaneously.
    Because the downward acceleration and the initial velocity in downward direction of the two bullets are the same, they will take the same time to hit the ground and for a half projectile.
    Time of flight = T =`` \sqrt{\frac{2h}{g}}``
  • Qstn #10
    The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be
    -(a) 25 m
    (b) 37 m
    (c) 50 m
    (d) 100 m
    digAnsr:   d
    Ans : (d) 100 m
    For the same u range, `` R\propto \,\mathrm{\,sin \,}\left(2\theta \right)``.
    So,
    `` \frac{{R}_{1}}{{R}_{2}}=\frac{\,\mathrm{\,sin \,}\left(2{\theta }_{1}\right)}{\,\mathrm{\,sin \,}\left(2{\theta }_{2}\right)}``
    `` \Rightarrow {R}_{2}=50\times \frac{\,\mathrm{\,sin \,}\left(90\right)}{\,\mathrm{\,sin \,}\left(30\right)}=100\,\mathrm{\,m \,}``
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