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.
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.
Explanation:
It is possible for an entity to move with a velocity other than zero when an external force that is net zero and unbalanced is given to the body. In actuality, once an object starts moving and a condition is enforced by any outside influence, the object will continue to move. The item must be moving in a specific direction and at a consistent speed.
2. When a carpet is beaten with a stick, dust comes out of it. Explain.
2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Explanation:
When a stick is used to strike a carpet, the energy applied by the stick causes the carpet to move. The carpet's inertia-filled dust particles prevent the carpet from changing its direction of motion. As a result, the carpet's forward momentum applies a backward pull to the dust particles, sending them moving in the other direction. Because of this, when the flooring is beaten, dust oozes out.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Explanation:
When some luggage is put on the top of a bus that is originally at rest, the bus's forward acceleration will cause the luggage to experience a force (in the opposite direction). A force (in the forward direction) is applied to the luggage when a bus that was originally in motion abruptly comes to rest due to the application of brakes. The baggage may tumble off the bus as a result of inertia, depending on its bulk and the power being applied. The baggage can be held in place and kept from sliding off the bus by tying it up.
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.
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.
Explanation:
The power of friction opposes the motion of the ball when it moves on a smooth surface of the earth. (The friction arises between the ground and the ball). The projectile is ultimately stopped by this frictional force. Consequently, the right response is (c).
The friction that develops between the ball and the flat ground will decrease if the surface is lubricated (with oil or another lubricant), allowing the ball to travel farther.
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.)
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.)
Explanation:
Given that the vehicle (s) traveled 400 meters,
The distance traveled in time (t) = 20 seconds.
The truck's starting speed (u) is equal to zero. (Since it starts from a state of rest)
Based on the velocity equation,
s = ut + ½ at2
400 = (0)20 + 1/2a (20)2
a = 2 m/s2
Mass of the truck M = 7 tonnes= 7000 kg (∵1 tonne = 1000 kg)
F = Ma =7000×2 = 14000 N
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?
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?
Explanation:
Given: Stone mass (m) = 1 kg
Initial speed (u) equals 20 m/s
0 m/s is the terminal speed (v). (The stone reaches a position of rest)
The stone (s) travelled for a distance of 50 m.
According to the third motion equation,
v² = u² + 2as
By changing the numbers in the aforementioned equation, we obtain,
0² = (20)² + 2(a) (50)
-400 = 100a
a = -400/100 = -4m/s² (retardation)
We know that
F = man
Substituting above obtained value of a = -4 in F = m x a
We get,
F = 1 × (-4) = -4N
Here, the opposite force, friction, is indicated by the negative symbol.
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
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
Explanation:
Given that (F) = 40,000 N is the force applied by the train
Frictional force = -5000 N (the negative sign indicates that the force is applied in the opposite direction)
Consequently, the overall increasing energy = 40,000 N + (-5000 N) = 35,000 N
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)
As a result, the train's acceleration is equal to its total accelerating power. (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?
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?
Explanation:
Considering the vehicle's Mass (m) = 1500 kg
Acceleration (a) = -1.7 ms-2
The second Law of Motion says, F = ma
F = 1500kg × (-1.7 ms-2) = -2550 N
Consequently, the force between the car and the road is -2550 N and is directed in the opposing direction of the car's velocity.
9. What is the momentum of an object of mass m, moving with a velocity v?
(mv)2 (b) mv2 (c) ½ mv2 (d) mv
9. What is the momentum of an object of mass m, moving with a velocity v?
(mv)2 (b) mv2 (c) ½ mv2 (d) mv
Explanation:
The option (D) mv is correct. The sum of the mass and the motion is known as the momentum.
Step 1: Illustrating motion expression
The momentum equation is expressed as p=mv.
Here, p denotes the momentum, m the mass, and v the object's motion.
An object's momentum is calculated by multiplying its mass (m) and speed (v).
Mass times speed equals momentum.
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?
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?
Explanation:
The cabinet's acceleration must be zero since its motion is constant. As a result, there is no effective power operating on it. This suggests that the 200 N force applied to the cabinet is equivalent to the amount of resisting frictional force. The overall contact force is 200 N as a result.
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?
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?
Explanation:
Given
First object's mass, m1, is 1.5 kg.
1.5 kg is the second object's mass.
First object's speed prior to impact, v1, is 2.5 m/s
The second object's speed, which is traveling in the opposing direction, is given by v2 = -2.5 m/s.
Having said that,
Pre-collision momentum equals post-collision momentum
m1v1 + m2v2 = (m1 + m2) v
1.5(2.5) + 1.5 (-2.5) = (1.5 + 1.5) v
3.75 – 3.75 = 3v
v = 0
As a result, the merged object's speed following the impact is 0 m/s.
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.
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.
Explanation:
The static friction between the vehicle and the road is very high because of the truck's extremely high bulk. The vehicle does not move when a tiny force is applied to propel it because the frictional force cancels out the applied force. Since the individual pushing the truck is not moved when the truck is stationary, it follows that the two forces are equivalent in strength but directed in the opposing direction. Thus, the student's reasoning is sound.
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.
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.
Explanation:
Given that the ball's mass (m) is 200g
The ball's initial speed (u) is 10 m/s.
The ball's final speed (v) is -5 m/s.
Ball's initial momentum = mu = 200g x 10 ms-1 = 2000 g.m.s-1
The ball's final velocity is calculated as follows: mv = 200g x -5 ms-1 = -1000 g.m.s-1
As a result, the shift in velocity (mv - mu) is equal to -1000 gm/s - 2000 gm/s - 3000 gm/s.
This suggests that when the hockey stick strikes the puck, the ball's velocity is reduced by 1000 g.m.s-1.
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.
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.
Explanation:
Given that the bullet's mass (m) is 10g (or 0.01 kg)
The bullet's initial speed (u) is 150 m/s.
Bullet's terminal speed is 0 meters per second (m/s).
Period (t) = 0.03 seconds
Calculating the bullet's motion is necessary to determine the impact distance.
Set the entry distance to s
The first rule of motion states that
v = u + at
0 = 150 + a (0.03)
a = -5000 ms-2
v2 = u2 + 2as
0 = 1502 + 2 x (-5000) s
s = 2.25 m
As per the second law of motion, F = ma
F = 0.01kg × (-5000 ms-2)
F = -50 N
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
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
Explanation:
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
The total momentum prior to the impact and the total momentum following the collision are identical, according to the rule of conservation of momentum. The overall momentum following the impact is thus also 10 kg.m.s-1.
Now, (m1 + m2) × v = 10kg.m.s-1
1.66 meters per second is the velocity of the final projectile.
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.
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.
Explanation:
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
The item therefore speeds at a rate of 0.5 ms-2. The force acting on the item (F = ma) is implied to be equivalent to:
50 N is equal to F = (100 kg) (0.5 ms-2)
The 100 kg item is therefore subjected to a 50 N force, which propels 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.
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.
Explanation:
The total momentum before the insect and the vehicle collided was identical to the total momentum after the collision, according to the rule of conservation of momentum. Consequently, the insect's change in motion is much larger than the car's change in momentum. (Since force is proportional to mass).
Part of Akhtar's presumption is accurate. The force applied to the insect during the impact is also very high due to the high mass of the vehicle. Kiran's assertion is untrue. Due to the principle of conservation of momentum, the shift in momentum of the insect and the vehicle is identical. Due to its mass, which is very low compared to the mass of a motorcar, an insect's velocity varies correspondingly. Similar to a bug, a motor vehicle has a very small motion due to its extremely high bulk.
Rahul is entirely correct in what he said. According to the third law of motion, the force that the bug applies to the car is equivalent to and in opposition to the force that the car applies to the insect. Rahul's assertion that the momentum shift is the same defies the momentum conservation rule, though.
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.
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.
Explanation:
Given that the dumbbell's mass (m) equals 10 kg
Covered distance (s): 80 centimeters = 0.8 meters
Initial speed (u) equals zero (it is dropped from a position of rest)
Speed of light (a) equals 10 ms-2
Terminal speed (v) =?
The dumbbell's impact momentum is equal to mv.
The third rule of motion states that
v2 – u2 = 2as
In light of this, v2 - 0 = 2 (10 ms-2) (0.8 m) = 16 m2s-2.
v = 4 m/s
The velocity that the dumbbell transmitted to the ground was equal to (10 kg) (4 m/s) = 40 kg.m.s-1.
