Chapter 9 Force And Laws Of Motion Solutions
Question - 11 : - Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer - 11 : -
When some luggage is placed on the roof of a bus which is initially at rest, the acceleration of the bus in the forward direction will exert a force (in the backward direction) on the luggage. In a similar manner, when a bus which is initially in a state of motion suddenly comes to rest due to the application of brakes, a force (in the forward direction) is exerted on the luggage.
Depending on the mass of the luggage and the magnitude of the force, the luggage may fall off the bus due to inertia. Tying up the luggage will secure its position and prevent it from falling off the bus.
Question - 12 : - 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 (a) the batsman did not hit the ball hard enough. (b) velocity is proportional to the force exerted on the ball. (c) there is a force on the ball opposing the motion. (d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Answer - 12 : -
When the ball rolls on the flat surface of the ground, its motion is opposed by the force of friction (the friction arises between the ground and the ball). This frictional force eventually stops the ball. Therefore, the correct answer is (c).
If the surface of the level ground is lubricated (with oil or some other lubricant), the friction that arises between the ball and the ground will reduce, which will enable the ball to roll for a longer distance.
Question - 13 : - 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 it’s mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Answer - 13 : -
Given, distance covered by the truck (s) = 400 meters
Time taken to cover the distance (t) = 20 seconds
Initial velocity of the truck (u) = 0 (since it starts from a state of rest)
Question - 14 : - A stone of 1 kg is thrown with a velocity of 20 ms-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?
Answer - 14 : -
Given, Mass of the stone (m) = 1kg
Initial velocity (u) = 20m/s
Terminal velocity (v) = 0 m/s (the stone reaches a position of rest)
Distance traveled by the stone (s) =
Question - 15 : - An 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 and (b) the acceleration of the train
Answer - 15 : -
(a) Given, forceexerted by the train (F) = 40,000 N
Force of friction =-5000 N (the negative sign indicates that the force is applied in the oppositedirection)
Therefore, the netaccelerating force = sum of all forces = 40,000 N + (-5000 N) = 35,000 N
(b) Total mass of thetrain = mass of engine + mass of each wagon = 8000kg + 5 × 2000kg
The total mass of thetrain is 18000 kg.
As per the second lawof motion, F = ma (or: a = F/m)
Therefore,acceleration of the train = (net accelerating force) / (total mass of thetrain)
= 35,000/18,000 = 1.94ms-2
The acceleration of thetrain is 1.94 m.s-2.
Question - 16 : - An automobile vehicle has a mass of 1500 kg.What must be the force between the vehicle and road if the vehicle is to bestopped with a negative acceleration of 1.7 ms-2?
Answer - 16 : -
Given, mass of thevehicle (m) = 1500 kg
Acceleration (a) =-1.7 ms-2
As per the second lawof motion, F = ma
F = 1500kg × (-1.7 ms-2)= -2550 N
Therefore, a force of2550 N must act on the vehicle in a direction opposite to that of its motion.
Question - 17 : - What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2 (b) mv2 (c) ½ mv2 (d)mv
Answer - 17 : -
Since momentum isdefined as the product of mass and velocity, the correct answer is (d), mv.
Question - 18 : - Using a horizontal force of 200 N, we intend to move a wooden cabinetacross a floor at a constant velocity. What is the friction force that will beexerted on the cabinet?
Answer - 18 : -
Since the velocity ofthe cabinet is constant, its acceleration must be zero. Therefore, theeffective force acting on it is also zero. This implies that the magnitude ofopposing frictional force is equal to the force exerted on the cabinet, whichis 200 N. Therefore, the total friction force is -200 N.
Question - 19 : - Two objects, each of mass 1.5 kg, are moving in the same straight linebut in opposite directions. The velocity of each object is 2.5 ms-1 beforethe collision during which they stick together. What will be the velocity ofthe combined object after collision?
Answer - 19 : -
Given, mass of theobjects (m1 and m2) = 1.5kg
Initial velocity ofthe first object (u1) = 2.5 m/s
Initial velocity ofthe second object which is moving in the opposite direction (u2) =-2.5 m/s
When the two massesstick together, the resulting object has a mass of 3 kg (m1 + m2)
Velocity of theresulting object (v) =?
As per the law ofconservation of momentum, the total momentum before the collision is equal tothe total momentum after the collision.
Total momentum beforethe collision = m1u1 + m2u2
= (1.5kg) (2.5 m/s) +(1.5 kg) (-2.5 m/s) = 0
Therefore, totalmomentum after collision = (m1+m2) v = (3kg) v = 0
Therefore v = 0
This implies that theobject formed after the collision has a velocity of 0 meters per second.
Question - 20 : - According to the third law of motion when we push on an object, theobject pushes back on us with an equal and opposite force. If the object is amassive truck parked along the roadside, it will probably not move. A studentjustifies this by answering that the two opposite and equal forces cancel eachother. Comment on this logic and explain why the truck does not move.
Answer - 20 : -
Since the truck has avery high mass, the static friction between the road and the truck is high. Whenpushing the truck with a small force, the frictional force cancels out theapplied force and the truck does not move. This implies that the two forces areequal in magnitude but opposite in direction (since the person pushing thetruck is not displaced when the truck doesn’t move). Therefore, the student’slogic is correct.