## Saturday, 6 July 2013

### CBSE Class IX (9th) Science | Chapter 9. FORCE AND LAWS OF MOTION | Solved Exercises

Question 1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Answer : Yes, it possible for an object to be traveling with a non-zero velocity, if the object experiences a net zero external unbalanced force.
This is due to inertia of motion. If the body is initially moving with some velocity, on a fine surface, then it will continue to move with the same velocity for some time, though the net external force acting on the body is zero. For example, when we stop pedaling a moving cycle, the bicycle continues to move for some time with a non-zero velocity, without application of a net zero external unbalanced force. After some time, the cycle slows down gradually and finally comes to rest due to force of friction and air resistance, acting in the opposite direction of motion. In the absence of unbalanced forces of ground friction and air resistance, acting in the opposite direction of motion, the bicycle will remain in motion for ever.

Question 2. When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer : When a carpet is beaten with a stick, its state of rest gets changed into state of motion. Where as the dust particles, which are loosely attached to its surface, try to remain in state of rest, due to static inertia, therefore get detached from the surface of carpet.

Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer : At the event of sudden change in speed or direction of the bus, when driver applies accelerator or brakes or steer it along the sharp curve on the road, the luggage kept on the roof tend to remain in its last position due to its state of inertia, which may be of either rest or of motion. This may result in, luggage being either falling in the backward direction or in forward direction or spinning off on the sharp road turns. Therefore,to prevent this, it advised to tie the luggage kept on the roof of a bus with a rope.

Question 4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
1. (a) the batsman did not hit the ball hard enough.
2. (b) velocity is proportional to the force exerted on the ball.
3. (c) there is a force on the ball opposing the motion.
4. (d) there is no unbalanced force on the ball, so the ball would want to come to rest.

Answer : (c) there is a force on the ball opposing the motion . Explanation : A cricket ball after getting hit by batsman, slows down and comes to rest due to forces of ground friction and air resistance acting in the opposite direction.

Question 5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)

 Given here Distance, s =400 m Initial Speed, s = 0 ms-1 Time, t = 20 s Acceleration a = ? As we know Distance, s = ut+ ½ (at2) Substituting the values in the above expression 400 m =0 ms-1× 20s + ½ a × (20s)2 => 400 m = a × 200 s2 => a = 400 m /200 s2 = 2 ms-2 Now we have to calculate the force by using the relation F = ma Here, m = 7 tonnes = 7 × 1000 kg, a = 2 ms-2 ∴ F = 7000 kg × 2 ms-2 = 14000 kg ms-2 =14000 N
So, the force acting on the truck is 14000 N

Question 6. A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

 Given here Mass (m) = 1 kg Initial Velocity (u) =20 ms-1 Total distance traveled (s) =50 m Force of friction, (F) = ? From the equation of motion, we have v2 − u2 = 2as => 2as =(v2 − u2) => 2a×50 m =02 − (20 ms-1)-2 => a = −400 m2s−2/ 100 m => = − 4ms−2 From Newton's second law of motion => Force (F) = mass(m) × acceleration(a) => F = 1 kg ×− 4ms−2 => = − 4 kg ms−2 = −4 N Ans Negative sign represents force of friction acting in opposite direction

CBSE Class IX (9th) Science | Chapter 9. FORCE AND LAWS OF MOTION |  Solved Exercises

Question 7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
1. (a) the net accelerating force;
2. (b) the acceleration of the train; and
3. (c) the force of wagon 1 on wagon 2.

 (a) Force exerted by the engine = 40,000 N Frictional Force offered by the track in opposite direction = 5,000 N The net accelerating force, F = 40,000 N − 5,000 N = 35,000 N (b) Mass of the train, m = 5 × 2,000kg + 8,000 kg =18,000 kg From Newton's second law of motion, acceleration a = F/m = 35,0000 N / 18,000 kg = 35 kg ms-2/ 18 kg = 1.94 ms-2 (c) The force exerted by wagon 1 on wagon 2 = mass of 4 wagons behind wagon 1 × acceleration = 8,000 kg × 1.94 ms−2 = 15,520 N Ans.

Question 8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms−2 ?

 As given Mass of vehicle, m = 1,500 kg Acceleration(Negative) = −1.7 ms−2 The force to be applied between the vehicle and road, F = ma = 1500 kg × (−1.7 ms−2 = −2,550 kg ms−2 = −2,550 N Ans

Question 9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2   (b) mv2   (c) mv2   (d) mv

Answer : (d) mv is correct as Momentum = mass×velocity = mv

Question 10.  Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

 To move wooden cabinet at a constant velocity, across the floor, the net force on cabinet should be zero ∴ Applied force + Frictional Force = Zero ∴Frictional Force + 200 N = 0 ∴ Frictional Force = −200 N Negative sign indicates the frictional force acts in the opposite direction of the applied forced. Hence, frictional force exerted on the cabinet will be 200 N

Question 11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

 As given, Mass of each object m1=m2 = 1.5 kg. Initial velocity of first object, u1 = 2.5 ms−1 Initial velocity of second object, u2 = 2.5 ms−1 Combined mass after the collision =m1 + m2 =1.5 kg + 1.5 kg = 3.0 kg Let v be the velocity of the combined object after collision By the law of conservation of momentum Total momentum before collision = Total momentum after collision m1u1+m2u2 =(m1+m2)v => 1.5 × 2.5 + 1.5 ×(&minus2.5) =(3.0)v => 0 = 3.0 v Or v = 0 ms−1 ∴ the combined object comes to rest after collision

Question 12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

Answer : The logic given by the student is wrong as the action and reaction forces act on different bodies.
Explanation for 'why the truck does not move' is as follows : The truck being a massive object has large static inertia, due to which, a large opposing force of friction comes into play between the ground and its tires, when ever an external force is applied to move it from its position. The force applied by us on the truck is insufficient to overcome the force of friction between its tires and road, rather it is counter balanced by a part of massive frictional force. Hence the truck does not move

Question 13. A hockey ball mass 200 g travelling at 10 m s−–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s−1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

 The Mass of Hockey ball, m = 200 g = 0.2 kg Initial velocity, u = 10 ms−1 Final velocity, v = −5ms−1 Initial momentum, p1 =mu = 0.2 ×10 = 2 kg ms−1 Final momentum, p2 =mu = 0.2 ×(−5) = &minus1 kg ms−1 ∴ Change in momentum =p2−p1 =&minus1 kg ms−1 − 2 kg ms−1 =− 3 kg ms−1 The negative sign shows a change in direction of hockey ball, along the return path after strike

Question 14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s−1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet

 Mass of bullet =10 g = 0.01 kg Initial Velocity of bullet, u = 150 ms−1 Final Velocity of bullet, v = 0 ms−1 Time taken, t =0.03 s a =(v−u)/t =(0 ms−1 &minus 150 ms−1)/0.03s = − 5000 ms−2 Let the distance of penetration = s As per equation of motion, v2 =u2+2as => s =(v2− u2)/2a = [0 − (150 ms−1)2]/2×(−5000ms−2) =(150×150 m2s−2)/2× 5000 ms−2 = 2.25 m ∴ s = 2.25 m The magnitude of the force exerted by the wooden block on the bullet, F =ma =0.01 kg ×(&minus 5000 ms−2 = 50 N

Question 15. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s−1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

 Here, mass of object m1, velocity of object, v1= 10 ms−1 Total momentum just before the impact = m1× v1 = 1 kg × 10 ms−1 = 10 kg ms−1 By the law of conservation of momentum Total momentum just after the impact =Total momentum just before the impact = 10 kg ms−1 Now, this object of mass 1 kg sticks to a stationary wooden block of mass 5 kg. ∴ combined mass of object, m2 = 1 kg + 5 kg = 6 kg Velocity of combined object, v2 = ? As per law of conservation of momentum m1v1 =m2v2 => v2 m1v1/m2 => v2 =(1 kg) ×(10 ms−1)/ 6 kg = 5/3 ms−1 ∴ The velocity of combined object is 5/3 ms−1

Question 16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–−1 to 8 ms−1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

 Mass of object, m = 100 kg Initial Velocity, ui = 5 ms−1 Final Velocity, v = 8 ms−1 Initial Momentum, p1 = mui= 500 kg ms−1 Final Momentum, p2 = mv= 800 kg ms−1 The magnitude of the force exerted on the object, F = (p2 - p1)/ t = (800 - 500) kg ms−1)/6 s = 50 kg ms−2 =300 kg ms−1/6 s = 50 N Ans.

Question 17. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

Answer : As the mass of insect is much smaller than that of motorcar and after collision, the velocity of insect will be same as that of motor car ; hence change in the momentum of insect can not be more than that of the motorcar. Hnece Kiran's suggestion is wrong
Akhtar's statement is also not right; as the force is not due to larger velocity, it is due to change in momentum.
Rahul's suggestion is correct. Its being a contained collision,  there is no external force on the system comprising insect and motor car; net momentum of system is conserved in the collision; hence
Δp1 + Δp2= 0
i.e. change in momentum of insect and motor car are equal and opposite. In other word action force exerted by insect on motor car is equal to reaction force exerted by motor car on insect.

Question 18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms−2 .

 Mass of dumb ball = 10 kg Height of fall, s =80 cm = 0.8 m Acceleration, a = 10 ms−2 Initial velocity, u = 0 Let the final velocity be = v As v2 = u2+ 2as => = 0 + 2 × 10 ms−2× 0.8 m => = 16 m2s−2 => v = 4 ms−1 The momentum transferred to the floor = mv = 10 kg ×4 ms−1 = 40 kg ms−1

CBSE Class IX (9th) Science | Chapter 9. FORCE AND LAWS OF MOTION | Solved Exercises

## Solved In text  Questions - Page :118 (With in chapter)

Question 1. Which of the following has more inertia:
(a) a rubber ball and a stone of the same size?
(b) a bicycle and a train?
(c) a five rupees coin and a one-rupee coin?

Answer. (a)Clearly, given the same size of both objects, the mass of the stone is much more than that of rubber ball. Therefor, stone has more inertia than rubber

(b) Here, as train has much larger mass than bicycle, therefor train has more inertia than bicycle

(c) Here, five rupee coin has more inertia as out of two coins, five rupee coin has larger mass than one rupee coin.

Question 2. In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”. Also identify the agent supplying the force in each case.

Answer. In the above example, the velocity of the ball changes three times. First change in velocity occurs, when A football player kicks a football to another player of his team. Second change in velocity takes place when another player kicks the football towards the goal. Third change takes place when The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team” :
All above changes in velocity of the ball, take place through two players and goalkeeper respectively as an agent supplying force.

Question 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch

Answer.When we shake a branch of tree vigorously, we set the branch structure into sudden motion. The leaves try to toretend theirs static position due to inertia of rest. which are not as strongly attached to tree, get detached from the branch of the tree, due to sudden impulsive force

Question 4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

Answer. When the bus is in motion, our speed is same as that of the bus.When a moving bus brakes to a stop, our lower part, being in contact with bus seat or floor, comes to rest, but the upper part, being not in immidiate contact with bus, tries to remain in state of motion in the forward direction, due to inertia of motion. Therefor, we fall in the forward direction.
Similiarly, when the bus accelerates from rest,our lower part, being in direct contact with bus seat or floor, comes into motion from the state of rest, where as the upper body tends to remain in state of rest due to static inertia; hence, we fall backward.

## Solved In text  Questions - Page :126 (With in chapter)

Chapter 9. FORCE AND LAWS OF MOTION | Science | CBSE Class IX (9th)

Question 1. If action is always equal to the reaction, explain how a horse can pull a cart.

Answer. As it appears, in pulling a cart, a horse exerts an equal amount of force in opposite direction to that exerted by the cart on the horse. Here forces being equal and opposite, should cancel out each other, and the cart should not move at all.
Which is, in fact, not the case, as we have seen, the cart does move, when a horse pulls it. Here the motion of cart is possible only, if horse pulls the cart with a force which is sufficient to overcome the force of friction between the cart tires and ground. When the forward reaction of horse pull on cart is greater than the backward reaction of cart pull on horse, the cart moves.

Question 2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Answer. When a fireman uses water jet from a hose pipe to dowse the fire, the water jet, due to large amount of water, and its high velocity, has a large forward momentum. This produces an equal amount of backward momentum in the hose pipe or its supporting structure. A fire man holding the hose pipe, has to apply a matching amount of force to offset this backward momentum generated as reaction from water jet of hose pipe. Hence, it is difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity

Question 3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s−–1 . Calculate the initial recoil velocity of the rifle.

 Mass of bullet, m1 = 50 g = 0.05 kg Mass of rifle, m2 = 4 kg Velocity of bullet, before firing, u1 = 0 Velocity of rifle, before firing, u2 = 0 Velocity of bullet, after firing, v1 = 35 m s−–1 Recoil velocity of bullet, after firing, v2 = ? Total momentum of bullet and rifle, before firing m1u1 + m2u2 = 0 Total momentum of bullet and rifle, after firing m1v1 + m2v2 = 0 =0,05 kg × 35 ms−1 + 4 kg ×v2 = 1.75 kg ms−1 + 4 kg×v2 We know, in the absence of an external force, as per the law of conservation of momentum Total momentum after firing = Total momentum before firing => = 1.75 kg ms−1 + 4 kg×v2 = 0 => v2 = −[(1.75 kg ms−1) / 4 kg ] => v2 = − 0.4375 ms−1 Ans. Negative sign shows the direction of recoil will be opposite to that of bullet

Question 4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s−–1 and 1 m s−–1, respectively.
They collide and after the collision, the first object moves at a velocity of 1.67 m s–1 . Determine the velocity of the second object.

 Mass of the first object, m1 = 100 g = 0.1 kg Mass of the second object, m2 = 200 g = 0.2 kg Before collision Velocity of the first object, u1 = 2 m s−–1 Velocity of the second object, u2 = 1 m s−–1 After collision Velocity of the first object, v1 = 1.67 m s−–1 Velocity of the second object, v2 = ? As per the law of conservation of momentum Total momentum before the collision = Total momentum after the collision m1u1 + m2u2 = m1v1 + m2v2 => 0.1 kg × 2 m s−–1 + 0,2 kg × 1 m s−–1 = 0.1 kg × 1.67 m s−–1 + 0.2 kg × v2 => 0.2kg ms−–1 + 0.2kg ms−–1 = 0.167 kg ms−–1 + 0.2 kg ×v2 => 0.4 kg ms−–1 − 0.167 kg ms−–1 = 0.2 kg ×v2 => v2 = 0.233 Kg ms−–1 / 0.2 kg => v2 = 1.165 ms−–1 Ans Hence, the velocity of second object is 1.165 ms−–1

CBSE Class IX (9th) Science | Chapter 9. FORCE AND LAWS OF MOTION | Solved Exercises

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