- PDF Free Download

Size: px

Start display at page:

Download ""

Kelley Owen
6 years ago
Views:

1 Laws of Motion. A fighter aircraft is looping in a vertical plane. The minimum velocity at the highest point is (Given r = radius of the loop) a) gr b) gr c) gr d) 3gr. In non-inertial frame, the second law of motion is written as a) F = ma b) F = ma + Fp c) F = ma Fp d) F = ma 3. A man of mass 6kg is riding in a lift. The weight of the man, when the lift is accelerating upwards and downwards at ms, are respectively (taking g = ms ) a) 7N and 48N b) 48N and 7N c) 6N and 6N d) None of these 4. The x and y-coordinates of a particle at any time t are given by x = 7t + 4t and y = 5y, where x and y are in metre and t is second. The acceleration of the particle at t = 5s is a) Zero b) 8ms c) ms d) 4ms 5. A ball of mass.5kg is moving with a velocity v of ms. It is subjected to a force of x Newton in s. Because of this force, the ball moves with a velocity of3ms. The value of x is a) 5N b) 8.5N c).5n d) N 6. A force F and 5N is required to push a car of mass kg slowly at constant speed on a leveled road. If a force F of N is applied, the acceleration of the car will be a) Zero b).5ms c) ms d).5ms

2 9 7. A monkey of mass m kg slides down a light rope attached to a fixed spring balance, with an acceleration a. The reading of this balance is w kg (g = acceleration due to gravity) a) wg m = b) m = w + a g a g c) The force of friction exerted by the rope on the monkey is m (g a) Newton d) The tension in the rope is wg Newton 8. A ball hits a vertical wall horizontally at m/s and bounces back at m/s, then 8 a) There is no acceleration because m/s -m/s = b) There may be an acceleration because its initial direction is horizontal c) There is an acceleration because there is a momentum change d) Even though there is no change in momentum there is a change in direction. Hence, it has acceleration 9. A body of mass kg has an initial velocity of 3ms along OE and it is subjected to a force of 4N in a direction perpendicular to OE. The distance of body from O after 4s will be a) m b) m c) 8m d) 48m. An object of mass kg moves at a constant speed ofms. A constant force, which acts for 4s on the object, gives it a speed ms in opposite direction. The force acting on the object is a) -3N b) -3N c) 3N d) 3N

3 7. A 6kg man stands on a spring scale in the lift. At some instant he finds, scale reading has changed from 6kg to 5kg for a while and then comes back to the original mark. What should we conclude? a) The lift was in constant motion upwards b) The lift while in motion downwards c) The lift while in constant motion upwards, is suddenly stopped d) The lift while in constant motion downwards, is suddenly stopped. A man weighs 8kg. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5ms. What would be the reading on the scale ( g = ms ) a) 8N b) N c) Zero d) 4N 3. When a car moves on a road with uniform speed of 3kmh, then resultant force on the car is 5 a) The driving force drives the car in the direction of propagation of car b) The resistive force acts opposite to the direction of propagation of car c) Zero d) None of the above 4. A person is standing in an elevator. In which situation he finds his weight less 4 a) When the elevator moves upward with constant acceleration b) When the elevator moves downward with constant acceleration c) When the elevator moves upward with uniform velocity d) When the elevator moves downward with uniform velocity 5. A ball of mass 5g moving with acceleration ms is hit by a force, which acts on it for.s. The impulsive force is a).5n-s b).n-s c).3n-s d).n-s

4 8 Conservation of Linear Momentum and Impulse 6. A particle of mass m is projected with velocity v making an angle of 45 with the horizontal. When the horizontal. When to particle lands on the level ground the magnitude of the change in its momentum will be a) mv b) mv / c) mv d) Zero 7. Sand is being dropped on a conveyor belt at the rate of Mkgs. The force necessary to keept the belt moving with a constant velocity of vms will be a) Mv Newton b) Mv Newton c) Mv Newton d) Zero 8. A machine gun is mounted on a kg vehicle on a horizontal smooth road (friction negligible). The gun fires bullets with a velocity of 5ms. If the mass of each bullet be g, what is the acceleration produced in the vehicle? 6 a) 5cms b) 3cms c) 5cms d) cms 9. A disc of mass g is kept floating horizontally in air by firing bullets, each of mass 5g with the same velocity at the same rate of bullets per second. The bullets rebound with the same speed in opposite direction, the velocity of each bullet at the time of impact is a) 96cms b) 9.8cms c) 98cms d) 98cms. A.5kg ball moving with a speed of ms strikes a hard wall at an angle of 3 with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for.5s, the average force acting on the wall is a) 48N b) 4N c) N d) 96N

5 5. In the figure given the position-time graph of a particle of mass.kg is shown. The impulse at t = s is a).kg ms b).kg ms c).kg ms d).4kg ms. A rocket of mass kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity ms with respect to the rocket. What is the minimum rate of burning of fuel, so as to just lift the rocket upwards against the gravitational attraction (take g = ms ) a) 5kgs b) kgs c) kgs d) 3. A force of N acts on a body of mass kg for s. The change in its momentum is 4 a) 5kg ms b) kg ms c) 3kg ms d) kg ms 4kgs 4. A boat of mass 4kg is at rest. A dog of mass 4kg moves in the boat with a velocity ofms. What is the velocity of boat? a) 4ms b) ms c) 8ms d) ms 5. A nucleus of mass under A, originally at rest, emits an α -particle with speed v. The daughter nucleus recoils with a speed a) v A + 4 4v c) A 4 b) d) 4v A + 4 v A 4 6. What is the momentum of a kg truck whose velocity is ms 5 a) kg ms 5 b) kg ms

6 c) 4kg ms d) None of these 7. Masses of two substances are g and 9g respectively. If their kinetic energies are same, then the ratio of their momentum will be 8 a) :9 b) 9: c) 3: d) :3 Equilibrium of Forces 8. Three forces acting on a body are shown in the figure. To have the resultant force only along the y-direction, the magnitude of the minimum additional force needed is a).5n b).5n c) 3 N d) 3N 4 9. The square of resultant of two equal forces is three times their product. Angle between the forces is 6 a) π b) π c) 4 π d) 3 π 3. A person used force (F), shown in figure to move a load with constant velocity on given surface. Identify the correct surface profile. a) b)

A) B) C) a) Pattern A is more sturdy b) Pattern B is more sturdy c) Pattern C is more sturdy d) All will have same sturdiness 3.

In order to make the rope perfectly horizontal, the force applied to each of its ends must be 4 a) Less than w b) Equal to w c) Equal to

7 c) d) 3. If a street light of mass M is suspend from the end of a uniform rod of length L in different possible pattern as shown in figure, then A) B) C) a) Pattern A is more sturdy b) Pattern B is more sturdy c) Pattern C is more sturdy d) All will have same sturdiness 3. A weight w is suspended from, the mid-point of a rope, whose ends are at the same level. In order to make the rope perfectly horizontal, the force applied to each of its ends must be 4 a) Less than w b) Equal to w c) Equal to w d) Infinitely large 33. Two equal forces (P each) act at a point inclined to each other at an angle of. The magnitude of their resultant is a) P/ b) P/4 c) P d) P Key ) c ) c 3) a 4) b 5) c 6) d 7) c 8) c 9) b ) b ) c ) b 3) c 4) b 5) d 6) c 7) a 8) a 9) d ) b ) a ) b 3) b 4) d 5) c 6) a 7) d 8) a 9) d 3) a 3) a 3) d 33) c

8 . mv mg = r v = gr Hints v = gr. In non-inertial frame, F = ma Fp Here F p is pseudo force and a is the acceleration of the body relative to non-inertial frame 3. Weight w = m (g+a) = 6(+) = 6x = 7 N 4. Acceleration 5. F t = m v a = a + a x y d x d y = + dt dt d x 8 = = dt F = x Newton, t = s, v = ms, v = ms, m =.5kg.5 x = =.5N 3 Net force 5 6. Acceleration = = =.5ms Mass 7. The reading of the spring balance = tension in the rope = force of friction between the rope and monkey = m (g) m (a) = m (g- a) Newton 8. Change in momentum = mv Also since we know that force = rate of change of momentum so, force will act on the ball. So there is acceleration. 9. The acceleration of the body perpendicular to OE is F a = m 4 = = ms Displacement along OE s = vt = 3 4 = m Displacement perpendicular to OE ms

9 (4) 6 s = at = = m The resultant displacement s = s + s = = m v u. a = t a = = 3ms 4 But,F = ma = x (-3) = -3N. Apparent weight = m ( g + a ) If lift suddenly stops during upward motion then apparent weight = m (g - a) As it is given that scale reading initially was 6 kg and due to sudden jerk reading decreases and finally comes back to the original mark i.e 6 kg. So, we can conclude that lift was moving upwards with constant speed and suddenly stops.. Mass of man M = 8 kg Acceleration of lift, a = 5ms When lift is moving upwards, the reading of weighting scale will be equal to R. The equation of motion gives R mg = ma Or R = mg + ma = m (g + a) = 8 ( +5) = N 3. Concept ' 4. w = m( g a) i.e, the person finds his weight less. 5. F= ma

10 m = 5g = 5 kg, a = ms F =.5 = 3N 3 Im pulse = F t = 3. =.3Ns Topic : Conversation of Linear Momentum and Impulse 6. Change in momentum, p p f pi = m v f vi = ( ) ( cos 45 sin 45 ) ( cos 45 sin 45 ) = m v i v j v i + v j v v v v = m i j i + j = p = ( ) d mv 7. Force required, F = dt dm = v vm dt As velocity v is constant, hence F = Mv Newton mv 8. Momentum carried by each bullet = mv =.x 5 Now, force = change in momentum in s = 5x = 5 N 5 Acceleration = ms = 5cms ' 9. From law of conversation of momentum., m vn = mg mg 98 v = = = 98cms ' m n 5. p = OB sin 3 ( OAsin 3 ) mvj kg ms = 5kg ms

11 ( mv ) = mv sin 3 sin 3 F t = mv sin 3 Or mv sin 3 F = t = mv sin 3 But, m=.5 kg, v = ms,t =.5 s, θ = 3.5 sin 3 F = = 4 N.5. I = change in momentum x = p = m v = m t x 4. But, m =. kg, = ms t 4 I =. =.kg ms. The velocity of exhaust gases with respect to rocket = ms The minimum force on the rocket to lift it Fmin = mg = = N dm Fmin Minimum rate of burning of fuel = = dt v = kgs 3. The change in momentum of the body p = F t = = kg ms 4. By law of conservation of linear momentum, mv = mv But, m = 4kg, m = 4kg, v = ms 4 4 v = 4 v = = ms A A X Y + α According to the law of conservation of momentum Or = myvy + mα vα

12 Or = ( A 4) v + 4v Or v y 4v = A 4 y Negative sign corresponds the recoil speed of daughter nucleus 6. Momentum of truck 7. By the following relation p = mk Given, m = g, m = 9g, K = K = K p p = = K 9 K 3 8. Minimum additional force needed F = F ( resultant ) x p = mv = = kg ms 5 Topic 3: Equilibrium of Forces Fresultant = [(4 )(cos 3 j sin 3 i) + (cos 6i + sin 6 j)] 3 3 i 3 3 = j + i + i + j = + j i F = = i F =.5N 9. P = Q and From, R = 3PQ = 3P R = P + Q + PQ θ cos 3P = P + P + P cosθ Or = 3P P P cosθ Or = cosθ cosθ =, thus, cosθ = cos 6 π cosθ =, thus cosθ = cos 6 or θ = 6 = 3 3. Concept 3. For equilibrium of street light, mg x = T y

13 mgx Or T = y For T to be minimum, y should be maximum Hence, pattern A is sturdier 3. For equilibrium of body mg = T cosθ mg T = cosθ For the string to be horizontal θ 9 mg T = T cos9 33. The resultant force of two equal forces = P + P + P. P cos P P P P = = =

PHYS 101 Previous Exam Problems. Force & Motion I

PHYS 101 Previous Exam Problems CHAPTER 5 Force & Motion I Newton s Laws Vertical motion Horizontal motion Mixed forces Contact forces Inclines General problems 1. A 5.0-kg block is lowered with a downward