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A train of mass M is moving on a circular track of radius 'R' with constant speed v. The length of the train is half of the perimeter of the track. The linear momentum of the train will be

$\frac{2 Mv}{π}$

MvR

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Explanation

$Angular velocity of the train, ω = \frac{v}{R} Linear velocity of centre of mass = ω \times \frac{2 R}{π} = \frac{2 v}{π} Momentum = M \times v = \frac{2 Mv}{π}$

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Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. Then the moment of inertia of the system about an axis about one of the sides of the square is

$M a^{2} + 2 M b^{2}$

$M a^{2}$

$M a^{2} + 4 M b^{2}$

$\frac{8}{5} M a^{2} + 2 M b^{2}$

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Explanation

$I = \frac{2}{5} M a^{2} \times 4 + M b^{2} \times 2 I = \frac{8}{5} M a^{2} + 2 M b^{2}$

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A body of mass (m) elastically collides with another stationary body of mass 3m. The fractional loss of K.E. of the body is:

$\frac{8}{9}$

$\frac{3}{4}$

$\frac{3}{5}$

$\frac{4}{7}$

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Explanation

$C o n s e r v e m o m e n t u m : m u = m v_{1} + 3 m v_{2} \Rightarrow 3 v_{2} + v_{1} = u (1) F o r e l a s t i c c o l l i s i o n : e = 1 = \frac{v_{2} - v_{1}}{u} \Rightarrow v_{2} - v_{1} = u (2) (1) + (2) g i v e s v_{2} = \frac{u}{2} v_{1} = - \frac{u}{2} \frac{∆ K}{K_{i}} = \frac{K_{i} - K_{f}}{K_{i}} = 1 - \frac{K_{f}}{K_{i}} = 1 - \frac{\frac{1}{2} m \frac{u^{2}}{4}}{\frac{1}{2} m u^{2}} = \frac{3}{4}$

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A hollow sphere is rolling without slipping on the inclined plane. The minimum coefficient of friction is :

$\frac{2}{5} \tan θ$

$\frac{2}{3} \tan θ$

$\frac{3}{4} \tan θ$

$\frac{5}{7} \tan θ$

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Explanation

$μ = \frac{\tan θ}{1 + \frac{R^{2}}{k^{2}}} w h e r e k i s r a d i u s o f g y r a t i o n F o r h o l l o w s p h e r e, \frac{R^{2}}{k^{2}} = \frac{3}{2} μ = \frac{\tan θ}{1 + \frac{3}{2}} = \frac{2 \tan θ}{5}$

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A body of mass (m) is projected with velocity v₀ at an angle of projection $θ$ . The magnitude of angular momentum about origin after time t from projection is :

Increase with time

decrease with time

independent of time

none of these

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Explanation

$τ = m g u \cos θ t \frac{d L}{d t} = m g u \cos θ t L = m g u \frac{\cos θ t^{2}}{2}$

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A body of mass m slides down an incline and reaches the bottom with a velocity v. If the same mass were in the form of a ring which rolls down this incline, the velocity of the ring at bottom would have been

$\sqrt{2} v$

$(1 / \sqrt{2}) v$

$(\sqrt{2 / 5}) v$

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Explanation

$v' = \sqrt{\frac{2 g h}{1 + \frac{K^{2}}{R^{2}}}}, v = \sqrt{2 g h} v' = \sqrt{\frac{2 g h}{1 + 1}} = \frac{\sqrt{2 g h}}{\sqrt{2}} = \frac{v}{\sqrt{2}}$

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A 100 g iron ball having velocity 10 m/s collides with a wall at an angle 30° and rebounds with the same angle. If the period of contact between the ball and wall is 0.1 second, then the force experienced by the wall is

10 N

100 N

1.0 N

0.1 N

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A man fires a bullet of mass 200 g at a speed of 5 m/s. The gun is of one kg mass. By what velocity the gun rebounds backwards ?

0.1 m/s

10 m/s

1 m/s

0.01 m/s

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Explanation

$v_{G} = \frac{m_{B} v_{B}}{m_{G}} = \frac{0 .2 \times 5}{1} = 1 m / s$

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A body of mass M at rest explodes into three pieces, two of which of mass M/4 each are thrown off in perpendicular directions with velocities of 3 m/s and 4 m/s respectively. The third piece will be thrown off with a velocity of

1.5 m/s

2.0 m/s

2.5 m/s

3.0 m/s

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Explanation

Momentum of one piece $= \frac{M}{4} \times 3$

Momentum of the other piece $= \frac{M}{4} \times 4$

∴ Resultant momentum $= \sqrt{\frac{9 M^{2}}{16} + M^{2}} = \frac{5 M}{4}$

The third piece should also have the same momentum. Let its velocity be v, then

$\frac{5 M}{4} = \frac{M}{2} \times v$ or $v = \frac{5}{2} = 2 .5 m / s e c$

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A force of 50 dynes is acted on a body of mass 5 g which acts for an interval of 3 seconds, then impulse is

0.15 × 10^–3 N-s

0.98 × 10^–3 N-s

1.5 × 10^–3 N-s

2.5 × 10^–3 N-s

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Explanation

Impulse = Force × Time = 50 × 10^–5 × 3

= 1.5 × 10^–3 N-s

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