Exercises Page: 128,129 ( Force And Laws Of Motion )
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.
Yes, it is possible. An object moving in some direction with constant velocity will continue in its state of motion as long as there are no external unbalanced forces acting on it. In order to change the motion of the object, some external unbalanced force must act upon it.
2. When a carpet is beaten with a stick, dust comes out of it. Explain.
When the carpet is beaten with a stick, the stick exerts a force on the carpet which sets it in motion. The inertia of the dust particles residing on the carpet resists the change in the motion of the carpet. Therefore, the forward motion of the carpet exerts a backward force on the dust particles, setting them in motion in the opposite direction. This is why the dust comes out of the carpet when beaten.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
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.
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 (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.
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.
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 it’s mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
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)
6. 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?
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) = 50 m
7. 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
(a) Given, force exerted by the train (F) = 40,000 N
Force of friction = -5000 N (the negative sign indicates that the force is applied in the opposite direction)
Therefore, the net accelerating force = sum of all forces = 40,000 N + (-5000 N) = 35,000 N
(b) Total mass of the train = mass of engine + mass of each wagon = 8000kg + 5 × 2000kg
The total mass of the train is 18000 kg.
As per the second law of motion, F = ma (or: a = F/m)
Therefore, acceleration of the train = (net accelerating force) / (total mass of the train)
= 35,000/18,000 = 1.94 ms-2
The acceleration of the train is 1.94 m.s-2.
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?
Given, mass of the vehicle (m) = 1500 kg
Acceleration (a) = -1.7 ms-2
As per the second law of motion, F = ma
F = 1500kg × (-1.7 ms-2) = -2550 N
Therefore, a force of 2550 N must act on the vehicle in a direction opposite to that of its motion.
9. 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
Since momentum is defined as the product of mass and velocity, the correct answer is (d), mv.
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?
Since the velocity of the cabinet is constant, its acceleration must be zero. Therefore, the effective force acting on it is also zero. This implies that the magnitude of opposing frictional force is equal to the force exerted on the cabinet, which is 200 N. Therefore, the total friction force is -200 N.
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 ms-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Given, mass of the objects (m1 and m2) = 1.5kg
Initial velocity of the first object (u1) = 2.5 m/s
Initial velocity of the second object which is moving in the opposite direction (u2) = -2.5 m/s
When the two masses stick together, the resulting object has a mass of 3 kg (m1 + m2)
Velocity of the resulting object (v) =?
As per the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Total momentum before the collision = m1u1 + m2u2
= (1.5kg) (2.5 m/s) + (1.5 kg) (-2.5 m/s) = 0
Therefore, total momentum after collision = (m1+m2) v = (3kg) v = 0
Therefore v = 0
This implies that the object formed after the collision has a velocity of 0 meters per second.
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.
Since the truck has a very high mass, the static friction between the road and the truck is high. When pushing the truck with a small force, the frictional force cancels out the applied force and the truck does not move. This implies that the two forces are equal in magnitude but opposite in direction (since the person pushing the truck is not displaced when the truck doesn’t move). Therefore, the student’s logic is correct.
13. A hockey ball of mass 200 g travelling at 10 ms–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms–1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Given, mass of the ball (m) = 200g
Initial velocity of the ball (u) = 10 m/s
Final velocity of the ball (v) = – 5m/s
Initial momentum of the ball = mu = 200g × 10 ms-1 = 2000 g.m.s-1
Final momentum of the ball = mv = 200g × –5 ms-1 = –1000 g.m.s-1
Therefore, the change in momentum (mv – mu) = –1000 g.m.s-1 – 2000 g.m.s-1 = –3000 g.m.s-1
This implies that the momentum of the ball reduces by 1000 g.m.s-1 after being struck by the hockey stick.
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.
Given, mass of the bullet (m) = 10g (or 0.01 kg)
Initial velocity of the bullet (u) = 150 m/s
Terminal velocity of the bullet (v) = 0 m/s
Time period (t) = 0.03 s
To find the distance of penetration, the acceleration of the bullet must be calculated.
Therefore, force exerted by the wooden block on the bullet (F) = 0.01kg × (-5000 ms-2)
= -50 N
This implies that the wooden block exerts a force of magnitude 50 N on the bullet in the direction that is opposite to the trajectory of the bullet.
15. An object of mass 1 kg travelling in a straight line with a velocity of 10 ms–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.
Given, mass of the object (m1) = 1kg
Mass of the block (m2) = 5kg
Initial velocity of the object (u1) = 10 m/s
Initial velocity of the block (u2) = 0
Mass of the resulting object = m1 + m2 = 6kg
Velocity of the resulting object (v) =?
Total momentum before the collision = m1u1 + m2u2 = (1kg) × (10m/s) + 0 = 10 kg.m.s-1
As per the law of conservation of momentum, the total momentum before the collision is equal to the total momentum post the collision. Therefore, the total momentum post the collision is also 10 kg.m.s-1
Now, (m1 + m2) × v = 10kg.m.s-1
The resulting object moves with a velocity of 1.66 meters per second.
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.
Given, mass of the object (m) = 100kg
Initial velocity (u) = 5 m/s
Terminal velocity (v) = 8 m/s
Time period (t) = 6s
Now, initial momentum (m × u) = 100kg × 5m/s = 500 kg.m.s-1
Final momentum (m × v) = 100kg × 8m/s = 800 kg.m.s-1
Therefore, the object accelerates at 0.5 ms-2. This implies that the force acting on the object (F = ma) is equal to:
F = (100kg) × (0.5 ms-2) = 50 N
Therefore, a force of 50 N is applied on the 100kg object, which accelerates it by 0.5 ms-2.
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.
As per the law of conservation of momentum, the total momentum before the collision between the insect and the car is equal to the total momentum after the collision. Therefore, the change in the momentum of the insect is much greater than the change in momentum of the car (since force is proportional to mass).
Akhtar’s assumption is partially right. Since the mass of the car is very high, the force exerted on the insect during the collision is also very high.
Kiran’s statement is false. The change in momentum of the insect and the motorcar is equal by conservation of momentum. The velocity of insect changes accordingly due to its mass as it is very small compared to the motorcar. Similarly, the velocity of motorcar is very insignificant because its mass is very large compared to the insect.
Rahul’s statement is completely right. As per the third law of motion, the force exerted by the insect on the car is equal and opposite to the force exerted by the car on the insect. However, Rahul’s suggestion that the change in the momentum is the same contradicts the law of conservation of momentum.
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.
Given, mass of the dumb-bell (m) = 10kg
Distance covered (s) = 80cm = 0.8m
Initial velocity (u) = 0 (it is dropped from a position of rest)
Acceleration (a) = 10ms-2
Terminal velocity (v) =?
Momentum of the dumb-bell when it hits the ground = mv
v = 4 m/s
The momentum transferred by the dumb-bell to the floor = (10kg) × (4 m/s) = 40 kg.m.s-1
Er. Aman Blogger, Educator