**Q.1 Â Â 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.
**

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

* Sol.*Â Â Â Beating of a carpet with a stick; makes the carpet come in motion suddenly, while dust particles trapped within the pores of carpet have tendency to remain in rest, and in order to maintain the position of rest they come out of carpet. This happens because of the application of Newtonâ€™s First Law of Motion which states that any object remains in its state unless any external force is applied over it.

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

* Sol.*Â Â Â Luggage kept on the roof of a bus has the tendency to maintain its state of rest when bus is in rest and to maintain the state of motion when bus is in motion according to Newtonâ€™s First Law of Motion.Â When bus will come in motion from its state of rest, in order to maintain the position of rest, luggage kept over its roof may fall down. Similarly, when a moving bus will come in the state of rest or there is any sudden change in velocity because of applying of brake, luggage may fall down because of its tendency to remain in the state of motion.Â This is the cause that it is advised to tie any luggage kept on the roof a bus with a rope so that luggage can be prevented from falling down.

**Q.4 Â Â 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
**

**Q.5 Â Â 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.)
**

Â Â Â Â Â Â Time (t) = 20Â Acceleration (a) = ?

Â Â Â Â Â Â We know that,

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Force acting upon truck:

Â Â Â Â Â Â Given mass of truck = 7 ton = 7 X 1000 kg = 7000 kg

Â Â Â Â Â Â We know that, force, P = m x a

Â Â Â Â Â Â Therefore, P = 7000 kg x 2 m s - 2

Â Â Â Â Â Â Or, P = 14000 Newton

Â Â Â Â Â Â Thus, Acceleration = 2 m s - 2 and force acting upon truck in the given condition = 14000 N

**Q.6 Â Â A stone of 1 kg is thrown with a velocity of 20 m/s 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? **

Â Â Â Â Â Â Mass of stone = 1 kg

Â Â Â Â Â Â Initial velocity, u = 20 m/s

Â Â Â Â Â Â Final velocity, v = 0 (as stone comes to rest)

Â Â Â Â Â Â Distance covered, s = 50 m

Â Â Â Â Â Â Force of friction = ?

Â Â Â Â Â Â We know that,

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Now, we know that, force, F = mass x acceleration

Â Â Â Â Â Â Therefore, F = 1 kg X -4ms-2

Â Â Â Â Â Â Or, F = -4ms-2

Â Â Â Â Â Â Thus, force of friction acting upon stone = -4ms-2. Here negative sign shows that force is being applied in the opposite direction of the movement of the stone.

**Q.7 Â Â 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:(a) the net accelerating force;(b) the acceleration of the train; and(c) the force of wagon 1 on wagon 2.
**

Â Â Â Â Â Â force of engine = 40000 NForce of friction = 5000 N

Â Â Â Â Â Â Mass of engine = 8000 kg

Â Â Â Â Â Â Total weight of wagons = 5 x 2000 kg = 10000 kg

Â Â Â Â Â Â (a) The net acceleratingÂ force

Â Â Â Â Â Â = Force exerted by engine â€“Â Force of fricition

Â Â Â Â Â Â = 40000 N â€“ 5000 N = 35000 N

Â Â Â Â Â Â (b) The acceleration of the trainWe know that, F = mass x acceleration

Â Â Â Â Â Â Or, 35000 N = (mass of engine + mass )

Â Â Â Â Â Â Or Â of 5 wagons X a

Â Â Â Â Â Â Or, 35000 N = (8000 kg + 10000 kg) X a

Â Â Â Â Â Â Or, 35000N = 18000 kg X a

Â Â Â Â Â Â (c) The force of wagon 1 on wagon 2

Â Â Â Â Â Â Since, net accelerating force = 35000 N

Â Â Â Â Â Â Mass of all 5 wagons = 10000 kg

Â Â Â Â Â Â We know that, F = m x a

Â Â Â Â Â Â Therefore,

Â Â Â Â Â Â

Â Â Â Â Â Â Therefore, acceleration of wagons = 3.5Â

Â Â Â Â Â Â Thus, force wagon 1 on 2 = mass of four wagons Ã— acceleration

Â Â Â Â Â Â Or, F = 4 Ã— 2000 kg Ã— 3.5Â

Â Â Â Â Â Â Or, F = 8000 kg Ã— 3.5Â

Â Â Â Â Â Â Or, F = 28000N

Â Â Â Â Â Â Thus,Â (a) The net accelerating force = 35000N

Â Â Â Â Â Â (b) The acceleration of train = 1.944Â

Â Â Â Â Â Â (c) The force of wagon 1 on 2 = 280000NÂ 35000N = 10000 kg x a

**Q.8 Â Â 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 m s-2?
**

Â Â Â Â Â Â Mass of the vehicle, m = 1500 kg

Â Â Â Â Â Â Acceleration, a = - 1.7 m s -2

Â Â Â Â Â Â Force acting between the vehicle and road,F = ?Â

Â Â Â Â Â Â We know that, F = m x a

Â Â Â Â Â Â Therefore, F = 1500 kg X 1.7 m s-2

Â Â Â Â Â Â Or, F = - 2550 N

Â Â Â Â Â Â Thus, force between vehicle and road = - 2550 N. Negative sign shows that force is acting in the opposite direction of the vehicle.

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

Â Â Â Â Â Â Explanation:

Â Â Â Â Â Â Given, mass = m, velocity = v,therefore, momentum =?

Â Â Â Â Â Â We know that, momentum, P = mass x velocity

Â Â Â Â Â Â Therefore, P = mv

Â Â Â Â Â Â Thus, option (d) mv is correct

**Q.10 Â Â 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?
**

**Q.11 Â Â 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?
**

Â Â Â Â Â Â Â Explanation:

Â Â Â Â Â Â Â Given,

Â Â Â Â Â Â Â Mass of first object, m1 = 1.5 kg

Â Â Â Â Â Â Â Mass of second object, m2 = 1.5 kg

Â Â Â Â Â Â Â Initial velocity of one object, u1 = 2.5 m/s

Â Â Â Â Â Â Â Initial velocity of second object, u2 = -2.5 m/s (Since second object is moving in opposite direction)

Â Â Â Â Â Â Â Final velocity of both the objects, which stick after

Â Â Â Â Â Â Â We know that,

Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â collision, v = ?

Â Â Â Â Â Â Â Therefore, final velocity of both the objects after collision will be zero.

**Q.12 Â Â 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.
**

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

Â Â Â Â Â Â Â Â Mass of hockey ball, m = 200 g = 200/1000 kg = 0.2 kg

Â Â Â Â Â Â Â Â Initial velocity of hockey ball, u = 10 m/s

Â Â Â Â Â Â Â Â Final velocity of hockey ball, v = - 5 m/s (because direction becomes opposite)

Â Â Â Â Â Â Â Â Change in momentum =?

Â Â Â Â Â Â Â Â We know Â that,

Â Â Â Â Â Â Â Â Momentum = mass x velocity

Â Â Â Â Â Â Â Â Therefore, Momentum of ball before getting struck

Â Â Â Â Â Â Â Â = 0.2 kg x 10 m/s = 2 kg m/sMomentum of ball after getting struck = 0.2 kg x - 5m/s = - 1 kg m/s

Â Â Â Â Â Â Â Therefore, change in momentum = momentum before getting struck â€“ momentum after getting struck= 2 kg m/s â€“ (-1 kg m/s) = 2 kg m/s + 1 kg m/s = 3 kg m/s

Â Â Â Â Â Â Â Thus, change of momentum of ball after getting struck = 3 kg m/s

**Q.14 Â Â Â A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s 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.
**

Â Â Â Â Â Â Â Mass of bullet, m = 10 g = 10/1000 kg = 0.01 kg

Â Â Â Â Â Â Â Initial velocity of bullet, u = 150 m/s

Â Â Â Â Â Â Â Since bullet comes to rest, thus final velocity, v = 0

Â Â Â Â Â Â Â Time, t = 0.03 s

Â Â Â Â Â Â Â Distance of penetration, i.e. Distance, covered (s) = ?

Â Â Â Â Â Â Â Magnitude of force exerted by wooden block = ?

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â We know that,

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â s = 4.5m â€“ 2.25m

Â Â Â Â Â Â Â s = 2.25m

Â Â Â Â Â Â Â We know that,Magnitude of force exerted by wooden block

Â Â Â Â Â Â Â We know that, Force = mass x acceleration

Â Â Â Â Â Â Â Or, F = 0.01 kg x â€“ 5000 m s-2 = - 50 N

Â Â Â Â Â Â Â Therefore,

Â Â Â Â Â Â Â Penetration of bullet in wooden block = 2.25 m

Â Â Â Â Â Â Â Force exerted by wooden block on bullet = - 50 N. Here negative sign shows that force is exerted in the opposite direction of bullet.

**Q.15 Â Â An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s 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.
**

Â Â Â Â Â Â Â Mass of the wooden block, m2 = 5kg

Â Â Â Â Â Â Â Initial velocity of object, u1 = 10 m/s

Â Â Â Â Â Â Â Initial velocity of wooden block, u2 = 0

Â Â Â Â Â Â Â Final velocity or moving object and wooden block, v = ?

Â Â Â Â Â Â Â Total momentum before collision and after collision = ?

Â Â Â Â Â Â Â We know that,

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Total momentum fo object and wooden block just before collision

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Total momentum just after collision

Â Â Â Â Â Â Â Â (Since both the objects move with same velocity 'v' after collision)

Â Â Â Â Â Â Â (From equation (i))

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Thus,

Â Â Â Â Â Â Â Velocity of both the object after collision = 1.66 m/s

Â Â Â Â Â Â Â Total momentum before collision = 10 kg m/s

Â Â Â Â Â Â Â Total momentum after collision = 10 kg m/s

**Q.16 Â Â An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.**

* Sol. Â Â Â *Given,

Â Â Â Â Â Â Â Initial velocity, u = 5 m/s

Â Â Â Â Â Â Â Final velocity, v = 8 m/s

Â Â Â Â Â Â Â Mass of the given object, m = 100 kg

Â Â Â Â Â Â Â Time, t = 6 s

Â Â Â Â Â Â Â Initial momentum and Final momentum = ?

Â Â Â Â Â Â Â Magnitude of force exerted on the object = ?

Â Â Â Â Â Â Â We know that,

Â Â Â Â Â Â Â Momentum = mass x velocity

Â Â Â Â Â Â Â Therefore, initial momentum = mass x initial velocity

Â Â Â Â Â Â Â = 100 kg X 5 m/s = 500 kg m/s

Â Â Â Â Â Â Â Final momentum = mass x final velocity

Â Â Â Â Â Â Â = 100 kg x 8 m/s = 800 kg m/s

Â Â Â Â Â Â Â We know that,

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Now, Force exerted on object = Mass x Acceleration

Â Â Â Â Â Â Â = 100 kg 0.5 m/s/s

Â Â Â Â Â Â Â = 50 N

**Q.17 Â Â 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.
**

**Q.18 Â Â 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 m/s-2.
**

Â Â Â Â Â Â Â Mass of dumb-bell = 10 kg

Â Â Â Â Â Â Â Distance, s = 80 cm = 80/100 = 0.8 m

Â Â Â Â Â Â Â Acceleration, a = 10 m/s/s

Â Â Â Â Â Â Â Initial velocity of dumb-bell, u = 0

Â Â Â Â Â Â Â Momentum = ?

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â We know that,Now, we know that, momentum = mass x velocity

Â Â Â Â Â Â Â = 10 kg x 4 m/s = 40 kg m/s

Â

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