Download Forces and Motion: A Guide to Contact, Non-Contact, Momentum, and Torque and more Study notes Physics in PDF only on Docsity!
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EXERCISE- 1 (A)
Question 1: What are contact forces? Give two Examples. Solution 1: The forces which act on bodies when they are in physical contact, are called contact forces. For e.g. frictional force and force exerted on two bodies during collision. Question 2: What are non – contact forces? Give two example. Solution 2: The forces experienced by bodies even without being physically touched, are called the non- contact forces. For e.g. Gravitational force and Electrostatic force. Question 3: Classify the following amongst contact and non-contact forces. (a) Frictional force (b) normal reaction force, (c) force of tension in a string (d) gravitation force (e) electrostatic force (f) magnetic force Solution 3: Contact force: (a) frictional force (b) normal reaction force (c) force of tension in a string Non-contact force: (d) gravitational force (e) electric force (f) magnetic force Question 4: Give one example in each case where: (a) the force is of contact and (b) Force is at a distance. Solution 4: (a) Force exerted on two bodies during collision. (b) Magnetic force between magnetic poles.
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Question 5: (a) A ball is hanging by a thread from the ceiling of the roof. Draw a neat labelled diagram showing the forces acting on the ball and the string. (b) A spring is compressed against a rigid wall. Draw a neat and labelled diagram showing the forces acting on the spring. Solution 5: Question 6: State one factor on which the magnitude of a non – contact force depends. How does it depend on the factor stated by you? Solution 6: The magnitude of a non-contact force depends on distance of separation of two bodies. Magnitude of force decreases as the distance increases.
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Question 11: State Newton’s first law of motion. Why is it called the law of inertia? Solution 11: Newton's first law of motion: A body continues to be in its state of rest or of uniform motion in a straight line unless an external force is applied on it. It is called the law of inertia because it tells that every material body has a property by virtue of which it resists the change in its state of rest or in its state of motion. This property is called inertia. Question 12: Define the term linear momentum. State its S.I. unit. Solution 12: The product of mass and velocity of the body is called linear momentum. S.I. unit of linear momentum is kg m s
- 1 . Question 13: (a) Write an expression for the change in momentum of a body of mass m moving with velocity v if (i) v ≪ c and (ii) v ⟶ c. (b) State the condition when the change in the momentum of a body depends only on the change in its velocity. Solution 13: (i) When v <c,. (ii) When v c, the change in momentum is. If velocity v of the moving body is much smaller than the velocity of light c (v<c) Question 14: How is force related to the momentum of a body? Solution 14: The rate of change of momentum of a body is directly proportional to the force applied on it and this change in momentum takes place in the direction of the applied force, i.e.,∝ ∆ ∆ ∆ ∆ Where p denotes momentum and ∆p is the change in momentum in time ∆t . Question 15: State Newton’s Second law of motion. Under what condition does it take the form F = ma? Solution 15:
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Newton's second law of motion - The rate of change of momentum of a body is directly proportional to the force applied on it and this change in momentum takes place in the direction of the applied force, i.e., F∝ ∆∆ where p denotes momentum and ∆p^ is the change in momentum in time ∆t. When mass m of the body is constant at velocity v (which is much smaller than the velocity of light c) then rate of change of momentum is: F =∆∆ = m ∆∆ = ma F= ma Question 16: Complete the following sentences: (a) Mass X change in velocity = ….. x time interval. (b) The mass of a body remains constant till the velocity of body is…… Solution 16: (a) Mass X change in velocity = Force x time interval. (b) The mass of a body remains constant till the velocity of body is Much less than the velocity of light. Question 17: Prove the force = mass x acceleration. State the condition when it holds. Solution 17: The rate of change of momentum = ∆ (^) = ∆ ( ) ∆ ∆ (When v c or m is not constant). But if mass m is constant i.e., v<c rate of change of momentum = Here the quantity ∆∆ = rate of change of velocity i.e., acceleration a. ∆^ ∆ = ∆^ ∆ Rate of change in momentum = ∆∆ = ∆∆ = Thus, by Newton's second law of motion, ∝ or, F= k ma Where k is a constant of proportionality which can be made equal to 1 by choosing the suitable unit for force. Hence, F = ma when mass m of the body is constant at velocity v which is much smaller than the velocity of light.
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(b) Yes, there is force acting on the rocket which is equal to the rate of change in momentum. Question 21: State Newton’s third law of motion. Solution 21: Third law of motion states that – To every action, there is always an equal and opposite reaction. Question 22: Name and define the S.I. and C.G.S. unit of force. How are they related? Solution 22: S.I. unit of force is Newton (N). One Newton is that force which when acting on a body of mass 1kg, produces an acceleration of 1 m s
- 2 in it. C.G.S. unit of force is dyne. One dyne is that force which when acting on a body of mass 1 gram, produces an acceleration of 1 cm s
- 2 in it. 1 Newton = 10 5 dyne Question 23: Define newton (the S.I. unit of force). Solution 23: S.I unit of force is newton (N). One Newton is that force which when acting on a body of mass 1kg, produces an acceleration of 1 m s
- 2 in it. Question 24: Define 1 kgf how is it related to newton? Solution 24: One kilogram force is the force by which the Earth pulls a mass of 1 kilogram towards itself. 1 kgf = 9.8 N Question 25: Explain what is understood by the following statement: 1 kilogram force (kgf) = 9.8 newton’
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Solution 25: One kilogram force is the force by which the Earth pulls a mass of 1 kilogram towards itself. 1 kgf = 9.8 N Question 26: How can you feel a force of 1 N? Solution 26: It means that we have to exert a force of 9.8 N to hold a mass of 1 kg on our palm. Question 27: Complete the following: (a) Force = mass x …….. (b) 1 N = ……… dyne (c) 1 N = ………… kgf (approx.) (d) newton is the unit of ………. Solution 27: (a) Force = mass x Acceleration (b) 1 N = 10 5 dyne (c) 1 N = 0.1 kgf (approx.) (d) Newton is the unit of Force. MULTIPLE CHOICE TYPE: Question 1:
- which of the following is not the force at a distance: (a) electrostatic force (b) gravitational force (c) frictional force (d) magnetic force. Solution 1: Frictional force Hint: It is the force which comes in play when two bodies come in contact.
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Question 3: Two balls of masses in ratio 1:2 are dropped from the same height find: (a) the ratio between their velocities when they strike the ground, and (b) the ratio of the forces acting on them during motion. Solution 3: (a) Initial velocity, u = 0 m/s Final velocity = v M 1 : M 2 =1 : 2 From equation: v^2 - u^2 = 2gh So, v = 1 √ 2 ℎ = √^2 ℎ^ = 1 = 1 ∶ 1 2 √^2 ℎ^1 (b) Force = mg (^1) = 1 = 1 = 1: 2 2 2 2 Question 4: A body X of mass 5 kg is moving with velocity 20 m s-^1 while another body Y of mass 20 kg is moving with velocity 5 m S
- 1 . Compare the momentum of the two bodies. Solution 4: Mass of body X = 5 kg Velocity of body X = 20 m/s Momentum of body X = mass x velocity P 1 = 5 × 20 =100 kgm/s Mass of body Y = 20kg Velocity of body Y = 5m/s Momentum of body Y = mass x velocity P 2 = 20 × 5 = 100 kgm/s (^1) = (^1 1) =^ 5×20 =^100 = 1: 1 20×5 100 2 2 2 Question 5: Calculate the acceleration produced in a body of mass 50g when acted upon by a force of 20 N. Solution 5: Mass= 50g= 0.05kg Force, F= 20N Force = mass x acceleration
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=ma a = = (^020) .05 = 400 −^2 Question 6: A car of mass 600 kg is moving with a speed of 10 ms-^1 while a scooter of mass 80kg is moving with a speed of 50 m s
- 1 (a) Compare their momentum (b) which vehicle will require more force to stop it in the (i) same interval of time (ii) same distance. Solution 6: Mass of the car, M = 600 kg Mass of the scooter, m = 80 kg Velocity of car, v = 10 m/s Velocity of Scooter = 50 m/s (a) we know that Momentum = mass × Velocity = = (600)(10) = 3 (80)(50) 2 Thus, the required ratio is 3: (b) (i) we know that force = ℎ ℎ Considering the time interval to be the same, Force a momentum The momentum of car (600 × 10 = 6000 kgm/s) Is more than the momentum of scooter (80 × 50 = 4000 kgm/s) The car will require more force to stop (ii) If we have to stop the vehicle at the same distance, Then the vehicle travelling with greater velocity will require more force to stop it. The velocity of scooter (= 50 m/s) is greater than the velocity of the car (= 10m/s), The scooter shall require more force to stop at the same distance. Question 7: How much acceleration will be produced in a body of mass 10 kg acted upon by a force of 2 kgf? (g = 9.8 ms
- 2 ) Solution 7: Mass of body = 10 kg Force = 2 kgf =2 × 9.8 =19. N Force = mass × acceleration = ma
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Question 10: A lead bullet of mass 20g, travelling with a velocity of 350 ms
- 1 , comes to rest after penetrating 40 cm in a still target. Find : (i) the resistive force offered by the target and (ii) the retardation caused by it. Solution 10: Mass of bullet=20g =0.020kg Initial velocity, u=350m/s Final velocity, v=0m/s Distance, s=40cm=0.4m From equation: v 2
- u 2 =2as 0 2
(350)^2 = 2a x 0. Acceleration a = − 350 × 2 ×0.4^ =^ −1.53 × 10^5 −^2 (i) Resistive force, F = mass x acceleration F= 0.020 × 1.53 × 10 5 = 3062.5N (ii) Retardation = - acceleration = 1.53 × 10^5 ms-^2 Question 11: A body of mass 50g is moving with a velocity of 10 ms
- 1 . It is brought to rest by a resistive force of 10 N. find: (i) the retardation and (ii) the distance that the body will travel after the resistive force is applied. Solution 11: Mass of body = 50 g = 0.050 kg Initial velocity, u = 10 m/s Final velocity, v = 0 m/s Resistive force, F = 10 N (i) Force, F = ma 10N = 0.050 × a Acceleration, a = 010 .05 = 200ms-^2 Retardation = − acceleration = 200 ms
- 2 (ii) From equation, v 2
- u 2 = 2as 0 2
- (10) 2 = 2 × (-200)× s −100 = − 400x s s = 100400 = 0.25 = 25
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Question 12: A uniform car of mass 500g travels with a uniform velocity of 25 m s
- 1 for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s calculate: (a) the retardation, (b) the distance which the car travels after the brakes are applied, (c) The force exerted by the brakes. Solution 12: Mass of toy car = 500 g = 0.5 kg Initial velocity, u = 25m/s Final velocity, v = 0m/s From equation: v = u + at 0 = 25 + a × 10 Acceleration, a = − 2.5ms-^2 (a) Retardation = − acceleration = 2.5ms-^2 (b) From equation : v^2 - u^2 = 2 as 0 2
- (25) 2 − 625 = −5s = 2 × (-2.5)× s S = 6255 = 125 (c) Force exerted by brake, F = ma F = 0.5 × (− 2.5) = −1.25 N Resistive force exerted by brakes = 1.25 N Question 13: A truck of mass 5 × 10 3 kg starting from rest travels a distance of 0.5 km in 10 s when a force is applied on it calculate: (a) the acceleration acquired by the truck and (b) the force applied Solution 13: Mass of truck = 5000 kg Initial velocity, u = 0 m/s Distance, s = 0.5 km = 500 m Time, t =10 s (a) From equation : S = ut + 500 = 0 × 10 + 12 12 2 × (10) 2 500 = 50a a = 50050 = 10 −^2 (b) Force = mass × acceleration
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Solution 2: The moment of force is equal to the product of the magnitude of the force and the perpendicular distance of the line of action of force from the axis of rotation. S.I. unit of moment of force is Newton metre (Nm). Question 3: Is moment of force a scalar or a vector? Solution 3: Vector Question 4: State two factor on which moment of force about a point depends. Solution 4: Moment of force about a point depends on the following two factors: (a)The magnitude of the force applied and, (b)The distance of line of action of the force from the axis of rotation. Question 5: When does a body rotate? State one way to change the direction of rotation of a body. Give a suitable example to explain your answer. Solution 5: When the body is pivoted at a point, the force applied on the body at a suitable point rotates the body about the axis passing through the pivoted point. The direction of rotation can be changed by changing the point of application of force. The given figure shows the anticlockwise and clockwise moments produced in a disc pivoted at its centre by changing the point of application of force F from A to B. Question 6: Write the expression for calculating the moment of force about a given a axis. Solution 6: Moment of force about a given axis = Force x perpendicular distance of force from the axis of rotation.
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Question 7: State one way to reduce the moment of given force about a given axis of rotation. Solution 7: Moment of force depends on the distance of line of action of the force from the axis of rotation. Decreasing the perpendicular distance from the axis reduces the moment of a given force. Question 8: What do you understand by the clockwise and anticlockwise moment of force? When is it taken positive? Solution 8: If the turning effect on the body is anticlockwise, moment of force is called anticlockwise moment and it is taken as positive while if the turning effect on the body is clockwise, moment of force is called clockwise moment and is taken negative. Question 9: Why is it easier to open a door by applying the force at the free end of it? Solution 9: It is easier to open a door by applying the force at the free end of it because larger the perpendicular distance, less is the force needed to turn the body. Question 10: The stone of hand flour grinder is provided with a handle near its rim. Give a reason. Solution 10: The stone of hand flour grinder is provided with a handle near its rim so that it can be rotated easily about the iron pivot at its centre by a small force applied at the handle. Question 11: It is easier to turn the steering wheel of a large diameter then that of a small diameter. Give reason. Solution 11: It is easier to turn the steering wheel of a large diameter than that of a small diameter because less force is applied on steering of large diameter which is at a large distance from the centre of rim.
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= − 4.4 Nm The negative sign suggests that the resultant torque is in the clockwise direction. Question 14: The adjacent diagram (Fig. 1.31) Shows a heavy roller, with its axle at O, which its axle at O, which is to be raised on a pavement XY by applying a minimum possible force. Show by an arrow on the diagram the point of application and the direction in which the force should be applied. Solution 14: Force F should be provided in the direction as shown in the diagram. Question 15: A body is acted upon by two forces each of magnitude F, but in opposite direction. State the effect of the forces if (a) both forces act at the same point of the body. (b) the two forces act at two different point of the body at a separation r. Solution 15: (a) Resultant force acting on the body = F − F = 0 moment of forces = 0 i.e., no motion of the body (b) The forces tend to rotate the body about the mid-point between two forces, Moment of forces = Fr Question 16: Draw a neat labelled diagram to show the direction of two forces acting on a body to produce rotation in it. Also mark the point about which rotation takes place, by the letter O.
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Solution 16: At A and B, two equal and opposite forces each of magnitude F are applied. The two forces rotate the bar in anticlockwise direction. Question 17: What do you understand by the term couple? State its effect. Give two examples of couple action in our daily life. Solution 17: Two equal and opposite parallel forces not acting along the same line, form a couple. A couple is always needed to produce the rotation. For example, turning a key in a lock and turning a steering wheel. Question 18: Define moment of couple. Write its S.I. unit. Solution 18: The moment of a couple is equal to the product of the either force and the perpendicular distance between the line of action of both the forces. S.I unit of moment of couple is Nm. Question 19: Prove that Moment of couple = Force × couple arm. Solution 19: At A and B, two equal and opposite forces each of magnitude F are applied. The two forces rotate the bar in anticlockwise direction. The perpendicular distance between two forces is AB
