Physics. TOPIC : Newton s law of Motion

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1 OPIC : ewton s law of Motion Date : Marks : 10 mks ime : ½ hr 1. A person sitting in an open car moving at constant velocity throws a ball vertically up into air. Outside the car In the car to the side of the person (b) In the car ahead of the person (d) Exactly in the hand which threw it up. For ordinary terrestrial experiments,, the observer in an inertial frame in the following cases is A child revolving in a giant wheel (b) A driver n a sports car moving with a constant high speed of 00 kmh -1 on a straight rod he pilot of an aeroplane which is taking off (d) A cyclist negotiating a sharp curve. he time period of a simple pendulum measured inside a stationary lift is found to be. If the lift starts accelerating upwards with an acceleration g/, the time period is (b) / 4. A vehicle of 100 kg is moving with a velocity of 5 m/sec. o stop it in 5000 (b) n small balls each of mass m impinge elastically each second on a surface with velocity u. he force experienced by the surface will be mnu (b) mnu 6. If a person with a spring balance and a body hanging from it goes up and up in an aeroplane, then the reading of the weight of the body as indicated by the spring balance will Go on increasing First increase and then decrease / (d) / (d) mnu (d) mnu (b) Go on decreasing (d) Remain the same 1 10 he ball falls sec, the required force in opposite direction is 7. A particle moves in the xy-plane under the action of a force F such that the components of its linear l momentum P at any time t are px cos t, py sin t. he angle between F and p at time t is 90 (b) (d) 0 8. he average force necessary to stop a bullet of mass 0 g moving with a speed of 50 m/s, as it penetrates into the wood for a distance of 1 cm is. 10 (b) (d) A man weighs 80 kg. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5m / s. What would be the reading on the scale. ( g 10m / s ) Page 1 of 8

2 400 (b) (d) Zero 10. A body of mass 1.0kg is falling with an acceleration of 10 m / sec. Its apparent weight will be ( g 10m / sec ) 1.0 kg wt (b).0kg wt 0.5 kg wt (d) Zero 11. A player caught a cricket ball of mass 150 gm moving at the rate of 0 m/sec. if the catching process be completed in 0.1 sec the force of the blow exerted by the ball on the hands of player is 0. (b) 0 00 (d) A student attempts to pull himself up by tugging on his hair. He will not succeed As the force exerted (b) the frictional force while gripping, is small ewton s law of inertia is not applicable to living being (d) As the force applied is internal to the system 1. A bird is sitting in a large closed cage which is placed on a spring balance. It records a weight of 5. he bird (mass m 0.5 kg) flies upward in the cage with an acceleration of m / s. he spring balance will now record a weight of 4 (b) 5 6 (d) he rate of mass of the gas emitted from rear of a rocket is initially 0.1 kg/sec. If the speed of the gas relative to the rocket is 50 m/sec and mass of the rocket is kg, then the acceleration of the rocket in m/sec is 5 (b) 5..5 (d) A man fires a bullet of mass 00 g at a speed of 5 m/s. he gun is of one kg mass. by what velocity the gun rebounds backwards 0.1 m/s (b) 10 m/s 1 m/s (d) 0.01 m/s 16. Rocket engines lift a rocket from the earth surface because hot gas with high velocity Push against the earth (b) Push against the air React against the rocket and push it up (d) Heat up the air which lifts the rocket 17. If a force of 50 act on body, the momentum acquired is 15 kg-m/s. What is the period for which force acts on the body 0.5 sec (b) 0. sec 0.4 sec (d) 0.5 sec 18. A diwali rocket is ejecting 0.05 kg of gases per second at a velocity of 400 m/sec. he accelerating force on the rocket is 0 dynes (b) 0 dynes (d) In a rocket of mass 1000 kg fuel is consumed at a rate of 40 kg/s. he velocity of the gases ejected from the rocket is 5 10 /. he thrust on the rocket is 10 (b) (d) A 5000 kg rocket is set for vertical firing. he exhaust speed is 800 ms 0 ms, the amount of gas ejected per second to supply the needed thrust will be ( g 10 ms Page of 8. o give an initial upward acceleration of )

17.5 kg s (b) 187.5 kg s 185.5 kg s (d) 17.5 kg s 1. he resultant of two forces P and P is R. If the first force is doubled then the resultant is also doubled.

3 17.5 kg s (b) kg s kg s (d) 17.5 kg s 1. he resultant of two forces P and P is R. If the first force is doubled then the resultant is also doubled. he angle between the two forces is o o 60 (b) 10 o 70 (d) o 180. he resultant of two forces, one double the other in magnitude, is perpendicular to the smaller of the two forces. he angle between the two forces is 0 60 (b) (d) 90. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block will be P (b) Pm M + m PM (d) M + m Pm M m 4. hree blocks of masses m 1, m and d m are connected by massless strings as shown on a frictionless table. hey are pulled with a force 40. If m 1 10 kg, m 6 kg and m 4 kg, the tension will be m1 1 m m 0 (b) (d) 5. wo rectangular blocks A and B of masses kg and kg respectively are connected by a spring of spring constant 10.8 m -1 and are placed on a frictionless horizontal surface. he block A was given an initial velocity of 0.15 ms -1 in the direction shown in the finre. he maximum compression of the spring during the motion is 0.15 ms -1 A B 0.01 m (b) 0.0 m 0.05 m (d) 0.0 m 6. A block of mass 4 kg is suspended through two light spring balances A and B. hen A and B will read respectively 4 kg and zero kg (b) Zero kg and 4 kg 4 kg and 4 kg A B 4kg P a g e

4 (d) kg and kg 7. wo blocks are connected by a string as shown in the diagram. he upper block is hung by another string. A force F applied on the upper string produces an acceleration of m / s in the upward direction in both the blocks. If and be the tensions in the two parts of the string, then (g 9.8 m/s ) and 47. (b) and 47. F kg and (d) and 0 ' 4 kg 8. When forces F 1, F, F are acting on a particle of mass m such that F and F are mutually perpendicular, then the particle remains stationary. If the force F 1 is now removed then the acceleration of the particle is F 1 / m (b) F F /mf1 ( F F )/ m (d) F / m 9. A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. he force on the pulley by the clamp is given by (b) Mg mg m (d) ( M + m) + m ( M + m) + M g g M 0. A particle of mass kg is initially at rest. A force acts on it whose magnitude chages with time.. he force time graph is shown below F() t (s) he velocity of the particle ager 10 s is 0 ms -1 (b)10 ms ms -1 (d) 6 ms -1 (e) 50 ms -1 Page 4 of 8

5 ASWERS 1. (d). (b). (b) (b) 1. (d) (b). (b) (b) (d) (d) (b) (b) (b) (d) 0. (e) Solutions 1. (d) Horizontal velocity of ball and person are same so both will cover equal horizontal distance in a given interval of time and after following the parabolic path the ball falls exactly in the hand which threw it up.. (b) l π and ' π g g 4 g [ As g' g + a g + ] ' l 4g/ dv F m 5000 dt Initially due to upward acceleration apparent weight of the body increases but then it decreases due to decrease in gravity. r 7. Given that p p ˆ xi + p ˆ y j cos t ˆi + sin t ˆj r r dp F sin t ˆi + cos t ˆj dt r ow, F. p r 0 i.e. angle between F r and p r is (d) u 50 m / s, v 0, s 0. 1 metre u v F ma m s F (50) For accelerated upward motion R m ( g + a) 80 (10 + 5) (d) R m ( g a) m (10 0) zero 11. (b) Force exerted by the ball dv 0 F m dt (d) As by an internal force momentumm of the system can not be changed. 1. Reading Weight of cage + Reaction by bird (10+) ,, Mass of the rocket kg. Mv constant / 5 P a g e

6 m BvB vg 1 m / s m 1 G 16. It works on the principle of conservation of momentum. 17. Change in momentum Impulse p 15 p F t t 0.5 sec F 50 dm 18. (b) F u dt dm hrust F u dt udm 0. (b) F m( g + a) dt dm m( g + a ) dt u 5000 (10 + 0) (b) R ( P) + (P) + P P cos θ (i) (R) (6 P) + (P) + 6P P cosθ (ii) by solving (i) and (ii), cosθ / θ 10 F sinθ. (b) tan α (as α 90 ) F + F cosθ. F + F cosθ 0 cos θ θ 10 1 M m F P kg / s θ R α 90 F P Acceleration of the system m + M he force exerted by rope on the masss 4. (d) ( m1 + m) m 1 + m + m 5. Block A moves with velocity 0.15ms -1 compresses the spring which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A. Let this velocity be c this state occurs when the spring is in a state of maximumm compression. Let x be the maximum compression in this stage. According to the law of conservation of linear momentum. we get 0.15 ms -1 MP m + M ( ) 40 0 A B mau(ma + mb)v Page 6 of 8

7 According to the conservation of energy, we get As the spring balances are massless therefore the reading of both balance should be equal. 7. FBD of mass kg FBD of mass 4kg 4 8 kg 4 kg (i) 40 8 By solving (i) and (ii) 47. and (ii) For equilibrium of system, F 1 F F + As θ 90 et force In the absence of force F 1, Acceleration Mass F + F F1 m m 9. (d) Force on the pulley by the clamp F pc + [( M + m) g] FPC m F pc ( Mg) + [( M + m) g] mg Mg F pc M + ( M + m) g 0. (e) Velocity after 10 sec. Area enclosed between F1 graph and time axis 7 P a g e

8 Area /. Page 8 of 8

A. B. C. D. E. v x. ΣF x

A. B. C. D. E. v x. ΣF x Q4.3 The graph to the right shows the velocity of an object as a function of time. Which of the graphs below best shows the net force versus time for this object? 0 v x t ΣF x ΣF x ΣF x ΣF x ΣF x 0 t 0