A meter scale is moving with uniform velocity. This implies:

1. the force acting on the scale is zero, but a torque about the centre of mass can act on the scale.
2. the force acting on the scale is zero and the torque acting about the centre of mass of the scale is also zero.
3. the total force acting on it need not zero but the torque on it is zero.
4. neither the force nor the torque needs to be zero.

Subtopic:  Newton's Laws | Application of Laws |
 60%
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A cricket ball of mass 150 g has an initial velocity \(\small {u = \left(3 \hat{i} + 4 \hat{j} \right) \text {ms}^{- 1}}\) and a  final velocity \(\small {v = - \left( 3 \hat{i} + 4 \hat{j} \right) \text{ms}^{- 1}}\), after being hit. The change in momentum (final momentum-initial momentum) is (in kgm/s)
1. \(\text {zero}\)
2. \(-\left ( 0.45\hat{i}+0.6\hat{j} \right ) \)
3. \(-\left ( 0.9\hat{i}+1.2\hat{j} \right ) \)
4. \(-5\left ( \hat{i} +\hat{j}\right ) \)

Subtopic:  Newton's Laws | Application of Laws |
 73%
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In the previous problem (3), the magnitude of the momentum transferred during the hit is:
1. zero
2. 0.75 kg-ms–1
3. 1.5 kg-ms–1
4. 14 kg-ms–1

Subtopic:  Newton's Laws | Application of Laws |
 68%
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Conservation of momentum in a collision between particles can be understood from:

1. Conservation of energy
2. Newton's first law only
3. Newton's second law only
4. Both Newton's second and third law
Subtopic:  Newton's Laws | Application of Laws |
 51%
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A hockey player is moving northward and suddenly turns westward at the same speed to avoid an opponent. The force that acts on the player is

1. frictional force along westward
2. muscle force along southward
3. frictional force along south-West
4. muscle force a south-West

Subtopic:  Newton's Laws | Application of Laws |
 67%
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A body of mass 2kg travels according to the law \(x \left( t \right) = pt + \left(qt\right)^{2} + \left(rt\right)^{3}\) where,\(\) \(p = 3 ~\text{ms }^{−1 },\) \(q = 4 ~\text{ms }^{−2}\) and \(r = 5 ~\text{ms }^{−3}\). The force acting on the body at \(t = 2 ~\text{s }\) is

1. 136 N 2. 134 N
3. 158 N 4. 68 N
Subtopic:  Newton's Laws | Application of Laws |
 66%
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A car of mass m starts from rest and acquires a velocity along the east, \(v=v\mathrm{\hat{i}}(v>0)\) in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is:

1. \(mv/2\) eastward and is exerted by the car engine.
2. \(mv/2\) eastward and is due to the friction on the tires exerted by the road.
3. more than \(mv/2\) eastward exerted due to the engine and overcomes the friction of the road.
4. \(mv/2\) exerted by the engine.

Subtopic:  Newton's Laws | Application of Laws |
 54%
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The motion of a particle of mass m is given by \(x=0\) for \(t<0 ~\text s\), \(x(t)=A~\text {sin}~4\pi t\) for \(0<t<(1/4) \text s ~(A>0)\), and \(x=0\) for \(t<(1/4) ~\text s\). Then:

(a) The force at \(t<(1/8) ~\text s\) on the particle is \(-16 \pi^2Am\)
(b) The particle is acted upon by an impulse of magnitude \(4 \pi Am\) at \(t=0 ~\text s\) and \(t<(1/4) ~\text s\)
(c) The particle is not acted upon by any force
(d) The particle is not acted upon by a constant force
(e) There is no impulse acting on the particle


Which of the following statement/s is/are true?

1. (a, c, d, e) 2. (a, c)
3. (b, c, d) 4. (a, b, d)
Subtopic:  Newton's Laws | Application of Laws |
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In the figure, the coefficient of friction between the floor and body \(B\) is \(0.1.\) The coefficient of friction between bodies \(B\) and \(A\) is \(0.2.\) A force \(F\) is applied as shown on \(B.\) The mass of \(A\) is \(rn/2\) and of \(B\) is \(m.\)

             

(a) The bodies will move together if \(F = 0.25~\text{mg}\)
(b) The \(A\) will slip with \(B\) if \(F = 0.5~\text{mg}\)
(c) The bodies will move together if \(F = 0.5~\text{mg}\)
(d) The bodies will be at rest if \(F = 0.1~\text{mg}\)
(e) The maximum value of \(F\) for which the two bodies will move together is \(0.45~\text{mg}\)


Which of the following statement(s) is/are true?

1. (a, b, d, e) 2. (a, c, d, e)
3. (b, c, d) 4. (a, b, c)
Subtopic:  Application of Laws | Friction |
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A body of mass \(10\) kg is acted upon by two perpendicular forces, \(6\) N and \(8\) N. The resultant acceleration of the body is:

(a) \(1~\text{ms}^{-2}\) at an angle of \(\text {tan}^{-1} \left(\dfrac{4}{3}\right ) \) w.r.t. \(6\) N force
(b) \(0.2~\text{ms}^{-2}\) at an angle of \(\text {tan}^{-1} \left(\dfrac{3}{4}\right ) \) w.r.t. \(8\) N force
(c) \(1~\text{ms}^{-2}\) at an angle of \(\text {tan}^{-1} \left(\dfrac{3}{4}\right ) \) w.r.t. \(8\) N force
(d) \(0.2~\text{ms}^{-2}\) at an angle of \(\text {tan}^{-1} \left(\dfrac{3}{4}\right ) \) w.r.t. \(6\) N force

Choose the correct option:
1. (a), (c) 2. (b), (c)
3. (c), (d) 4. (a), (b), (c)

Subtopic:  Application of Laws |
 78%
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