Mass of the automobile vehicle, m = 1500 kg Final velocity, v = 0 m/s Acceleration of the automobile, a = –1.7 ms⁻² From Newton's second law of motion: Force = Mass × Acceleration = 1500 × (–1.7) = –2550 N Hence, the force between the automobile and the road is –2550 N, in the direction opposite toRead more
Mass of the automobile vehicle, m = 1500 kg
Final velocity, v = 0 m/s
Acceleration of the automobile, a = –1.7 ms⁻²
From Newton’s second law of motion:
Force = Mass × Acceleration = 1500 × (–1.7) = –2550 N
Hence, the force between the automobile and the road is –2550 N, in the direction
opposite to the motion of the automobile.
(a) 35000 N (b) 1.944 m/s² (c) 15552 N (a) Force exerted by the engine, F = 40000 N Frictional force offered by the track, Ff = 5000 N Net accelerating force, Fa = F - Ff = 40000 - 5000 = 35000 N Hence, the net accelerating force is 35000 N. (b) Acceleration of the train = a The engine exerts a forcRead more
(a) 35000 N (b) 1.944 m/s² (c) 15552 N
(a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, Ff = 5000 N
Net accelerating force, Fa = F – Ff = 40000 – 5000 = 35000 N
Hence, the net accelerating force is 35000 N.
(b) Acceleration of the train = a
The engine exerts a force of 40000 N on all the five wagons.
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, m = Mass of a wagon × Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
∴ m = 2000 × 5 = 10000 kg
Total mass (including the mass of engine), M = m + 8000 = 18000 kg
Fa = Ma
⇒ 𝑎 = 𝐹𝑎/𝑀 = 35000/18000 = 1.944 𝑚/𝑠²
(c) Mass of 4 wagons (excluding wagon 1) m = 2000 × 4 = 8000 kg
Acceleration of the train = 1.944 𝑚/𝑠²
∴ The force of wagon 1 on wagon 2 = 𝑚𝑎 = 8000 × 1.944 = 15552 𝑁
Initial velocity of the stone, u = 20 m/s Final velocity of the stone, v = 0 m/s Distance covered by the stone, s = 50 m According to the third equation of motion: v² = u² + 2as Where, Acceleration, a (0)2 = (20)² + 2 × a × 50 a = – 4 m/s² The negative sign indicates that acceleration is acting agaiRead more
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0 m/s
Distance covered by the stone, s = 50 m
According to the third equation of motion:
v² = u² + 2as
Where,
Acceleration, a
(0)2 = (20)² + 2 × a × 50
a = – 4 m/s²
The negative sign indicates that acceleration is acting against the motion of the
stone.
Mass of the stone, m = 1 kg
From Newton’s second law of motion:
Force,
F = Mass x Acceleration
F= ma
F= 1 × (– 4) = – 4 N
Hence, the force of friction between the stone and the ice is – 4 N.
Initial velocity of the truck, u = 0 m/s Time taken, t = 20 s Distance covered by the stone, s = 400 m According to the second equation of motion: 𝑠 = 𝑢𝑡 +1/2 𝑎𝑡² We have 400 = 0 × 20 + 1/2× 𝑎 × (20)² ⇒ 400 = 200𝑎 ⇒ 𝑎 = 2 𝑚/𝑠 ² Now, Force = mass × acceleration ⇒ 𝐹 = 7000 × 2 = 14000 𝑁 For more answeRead more
Initial velocity of the truck, u = 0 m/s
Time taken, t = 20 s
Distance covered by the stone, s = 400 m
According to the second equation of motion:
𝑠 = 𝑢𝑡 +1/2 𝑎𝑡²
We have
400 = 0 × 20 + 1/2× 𝑎 × (20)²
⇒ 400 = 200𝑎
⇒ 𝑎 = 2 𝑚/𝑠 ²
Now, Force = mass × acceleration
⇒ 𝐹 = 7000 × 2 = 14000 𝑁
(c) A batsman hits a cricket ball, which then rolls on a level ground. After covering a short distance, the ball comes to rest because there is frictional force on the ball opposing its motion. Frictional force always acts in the direction opposite to the direction of motion. Hence, this force is reRead more
(c) A batsman hits a cricket ball, which then rolls on a level ground. After
covering a short distance, the ball comes to rest because there is frictional
force on the ball opposing its motion.
Frictional force always acts in the direction opposite to the direction of motion.
Hence, this force is responsible for stopping the cricket ball.
When the bus accelerates and moves forward, it acquires a state of motion. However, the luggage kept on the roof, owing to its inertia, tends to remain in its state of rest. Hence, with the forward movement of the bus, the luggage tends to remain at its original position and ultimately falls from thRead more
When the bus accelerates and moves forward, it acquires a state of motion.
However, the luggage kept on the roof, owing to its inertia, tends to remain in its
state of rest. Hence, with the forward movement of the bus, the luggage tends to
remain at its original position and ultimately falls from the roof of the bus. To
avoid this, it is advised to tie any luggage kept on the roof of a bus with a rope.
Inertia of an object tends to resist any change in its state of rest or state of motion. When a carpet is beaten with a stick, then the carpet comes to motion. But, the dust particles try to resist their state of rest. According to Newton's first law of motion, the dust particles stay in a state ofRead more
Inertia of an object tends to resist any change in its state of rest or state of motion.
When a carpet is beaten with a stick, then the carpet comes to motion. But, the
dust particles try to resist their state of rest. According to Newton’s first law of
motion, the dust particles stay in a state of rest, while the carpet moves. Hence,
the dust particles come out of the carpet.
Yes. Even when an object experiences a net zero external unbalanced force, it is possible that the object is travelling with a non-zero velocity. This is possible only when the object has been moving with a constant velocity in a particular direction. Then, there is no net unbalanced force applied oRead more
Yes. Even when an object experiences a net zero external unbalanced force, it is
possible that the object is travelling with a non-zero velocity. This is possible only
when the object has been moving with a constant velocity in a particular direction.
Then, there is no net unbalanced force applied on the body. The object will keep
moving with a non-zero velocity. To change the state of motion, a net non-zero
external unbalanced force must be applied on the object.
Mass of one of the objects, m₁ = 100 g = 0.1 kg Mass of the other object, m₂ = 200 g = 0.2 kg Velocity of m1 before collision, v₁ = 2 m/s Velocity of m2 before collision, v₂ = 1 m/s Velocity of m1 after collision, v₄ = 1.67 m/s Velocity of m2 after collision = v₄ According to the law of conservationRead more
Mass of one of the objects, m₁ = 100 g = 0.1 kg
Mass of the other object, m₂ = 200 g = 0.2 kg Velocity of m1 before collision, v₁ = 2 m/s
Velocity of m2 before collision, v₂ = 1 m/s
Velocity of m1 after collision, v₄ = 1.67 m/s
Velocity of m2 after collision = v₄
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄
⇒ 0.1×2 + 0.2×1 = 0.1×1.67 + 0.2×v₄
⇒ 0.4 = 0.67 + 0.2×v₄
⇒ v₄ = 1.165 m/s
Hence, the velocity of the second object becomes 1.165 m/s after the collision.
Mass of the rifle, m1 = 4 kg Mass of the bullet, m2 = 50g = 0.05 kg Recoil velocity of the rifle = v₁ Bullet is fired with an initial velocity, v₂ = 35 m/s Initially, the rifle is at rest. Thus, its initial velocity, v = 0 Total initial momentum of the rifle and bullet system = (m1 + m2) v = 0 TotalRead more
Mass of the rifle, m1 = 4 kg
Mass of the bullet, m2 = 50g = 0.05 kg
Recoil velocity of the rifle = v₁
Bullet is fired with an initial velocity, v₂ = 35 m/s
Initially, the rifle is at rest.
Thus, its initial velocity, v = 0
Total initial momentum of the rifle and bullet system = (m1 + m2) v = 0
Total momentum of the rifle and bullet system after firing:
= m₁ v₁ + m2v2 = 4(v1) + 0.05×35 = 4v1 + 1.75
According to the law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing
4 v₁ + 1.75 = 0
⇒ v₁ = – 1.75/4 = – 0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.
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?
Mass of the automobile vehicle, m = 1500 kg Final velocity, v = 0 m/s Acceleration of the automobile, a = –1.7 ms⁻² From Newton's second law of motion: Force = Mass × Acceleration = 1500 × (–1.7) = –2550 N Hence, the force between the automobile and the road is –2550 N, in the direction opposite toRead more
Mass of the automobile vehicle, m = 1500 kg
Final velocity, v = 0 m/s
Acceleration of the automobile, a = –1.7 ms⁻²
From Newton’s second law of motion:
Force = Mass × Acceleration = 1500 × (–1.7) = –2550 N
Hence, the force between the automobile and the road is –2550 N, in the direction
opposite to the motion of the automobile.
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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.
(a) 35000 N (b) 1.944 m/s² (c) 15552 N (a) Force exerted by the engine, F = 40000 N Frictional force offered by the track, Ff = 5000 N Net accelerating force, Fa = F - Ff = 40000 - 5000 = 35000 N Hence, the net accelerating force is 35000 N. (b) Acceleration of the train = a The engine exerts a forcRead more
(a) 35000 N (b) 1.944 m/s² (c) 15552 N
(a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, Ff = 5000 N
Net accelerating force, Fa = F – Ff = 40000 – 5000 = 35000 N
Hence, the net accelerating force is 35000 N.
(b) Acceleration of the train = a
The engine exerts a force of 40000 N on all the five wagons.
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, m = Mass of a wagon × Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
∴ m = 2000 × 5 = 10000 kg
Total mass (including the mass of engine), M = m + 8000 = 18000 kg
Fa = Ma
⇒ 𝑎 = 𝐹𝑎/𝑀 = 35000/18000 = 1.944 𝑚/𝑠²
(c) Mass of 4 wagons (excluding wagon 1) m = 2000 × 4 = 8000 kg
Acceleration of the train = 1.944 𝑚/𝑠²
∴ The force of wagon 1 on wagon 2 = 𝑚𝑎 = 8000 × 1.944 = 15552 𝑁
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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?
Initial velocity of the stone, u = 20 m/s Final velocity of the stone, v = 0 m/s Distance covered by the stone, s = 50 m According to the third equation of motion: v² = u² + 2as Where, Acceleration, a (0)2 = (20)² + 2 × a × 50 a = – 4 m/s² The negative sign indicates that acceleration is acting agaiRead more
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0 m/s
Distance covered by the stone, s = 50 m
According to the third equation of motion:
v² = u² + 2as
Where,
Acceleration, a
(0)2 = (20)² + 2 × a × 50
a = – 4 m/s²
The negative sign indicates that acceleration is acting against the motion of the
stone.
Mass of the stone, m = 1 kg
From Newton’s second law of motion:
Force,
F = Mass x Acceleration
F= ma
F= 1 × (– 4) = – 4 N
Hence, the force of friction between the stone and the ice is – 4 N.
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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 metric tonnes (Hint: 1 metric tonne = 1000 kg.)
Initial velocity of the truck, u = 0 m/s Time taken, t = 20 s Distance covered by the stone, s = 400 m According to the second equation of motion: 𝑠 = 𝑢𝑡 +1/2 𝑎𝑡² We have 400 = 0 × 20 + 1/2× 𝑎 × (20)² ⇒ 400 = 200𝑎 ⇒ 𝑎 = 2 𝑚/𝑠 ² Now, Force = mass × acceleration ⇒ 𝐹 = 7000 × 2 = 14000 𝑁 For more answeRead more
Initial velocity of the truck, u = 0 m/s
Time taken, t = 20 s
Distance covered by the stone, s = 400 m
According to the second equation of motion:
𝑠 = 𝑢𝑡 +1/2 𝑎𝑡²
We have
400 = 0 × 20 + 1/2× 𝑎 × (20)²
⇒ 400 = 200𝑎
⇒ 𝑎 = 2 𝑚/𝑠 ²
Now, Force = mass × acceleration
⇒ 𝐹 = 7000 × 2 = 14000 𝑁
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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
(c) A batsman hits a cricket ball, which then rolls on a level ground. After covering a short distance, the ball comes to rest because there is frictional force on the ball opposing its motion. Frictional force always acts in the direction opposite to the direction of motion. Hence, this force is reRead more
(c) A batsman hits a cricket ball, which then rolls on a level ground. After
covering a short distance, the ball comes to rest because there is frictional
force on the ball opposing its motion.
Frictional force always acts in the direction opposite to the direction of motion.
Hence, this force is responsible for stopping the cricket ball.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
When the bus accelerates and moves forward, it acquires a state of motion. However, the luggage kept on the roof, owing to its inertia, tends to remain in its state of rest. Hence, with the forward movement of the bus, the luggage tends to remain at its original position and ultimately falls from thRead more
When the bus accelerates and moves forward, it acquires a state of motion.
However, the luggage kept on the roof, owing to its inertia, tends to remain in its
state of rest. Hence, with the forward movement of the bus, the luggage tends to
remain at its original position and ultimately falls from the roof of the bus. To
avoid this, it is advised to tie any luggage kept on the roof of a bus with a rope.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
When a carpet is beaten with a stick, dust comes out of it. Explain.
Inertia of an object tends to resist any change in its state of rest or state of motion. When a carpet is beaten with a stick, then the carpet comes to motion. But, the dust particles try to resist their state of rest. According to Newton's first law of motion, the dust particles stay in a state ofRead more
Inertia of an object tends to resist any change in its state of rest or state of motion.
When a carpet is beaten with a stick, then the carpet comes to motion. But, the
dust particles try to resist their state of rest. According to Newton’s first law of
motion, the dust particles stay in a state of rest, while the carpet moves. Hence,
the dust particles come out of the carpet.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
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. Even when an object experiences a net zero external unbalanced force, it is possible that the object is travelling with a non-zero velocity. This is possible only when the object has been moving with a constant velocity in a particular direction. Then, there is no net unbalanced force applied oRead more
Yes. Even when an object experiences a net zero external unbalanced force, it is
possible that the object is travelling with a non-zero velocity. This is possible only
when the object has been moving with a constant velocity in a particular direction.
Then, there is no net unbalanced force applied on the body. The object will keep
moving with a non-zero velocity. To change the state of motion, a net non-zero
external unbalanced force must be applied on the object.
For more answers visit to website:
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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 one of the objects, m₁ = 100 g = 0.1 kg Mass of the other object, m₂ = 200 g = 0.2 kg Velocity of m1 before collision, v₁ = 2 m/s Velocity of m2 before collision, v₂ = 1 m/s Velocity of m1 after collision, v₄ = 1.67 m/s Velocity of m2 after collision = v₄ According to the law of conservationRead more
Mass of one of the objects, m₁ = 100 g = 0.1 kg
Mass of the other object, m₂ = 200 g = 0.2 kg Velocity of m1 before collision, v₁ = 2 m/s
Velocity of m2 before collision, v₂ = 1 m/s
Velocity of m1 after collision, v₄ = 1.67 m/s
Velocity of m2 after collision = v₄
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄
⇒ 0.1×2 + 0.2×1 = 0.1×1.67 + 0.2×v₄
⇒ 0.4 = 0.67 + 0.2×v₄
⇒ v₄ = 1.165 m/s
Hence, the velocity of the second object becomes 1.165 m/s after the collision.
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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 the rifle, m1 = 4 kg Mass of the bullet, m2 = 50g = 0.05 kg Recoil velocity of the rifle = v₁ Bullet is fired with an initial velocity, v₂ = 35 m/s Initially, the rifle is at rest. Thus, its initial velocity, v = 0 Total initial momentum of the rifle and bullet system = (m1 + m2) v = 0 TotalRead more
Mass of the rifle, m1 = 4 kg
Mass of the bullet, m2 = 50g = 0.05 kg
Recoil velocity of the rifle = v₁
Bullet is fired with an initial velocity, v₂ = 35 m/s
Initially, the rifle is at rest.
Thus, its initial velocity, v = 0
Total initial momentum of the rifle and bullet system = (m1 + m2) v = 0
Total momentum of the rifle and bullet system after firing:
= m₁ v₁ + m2v2 = 4(v1) + 0.05×35 = 4v1 + 1.75
According to the law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing
4 v₁ + 1.75 = 0
⇒ v₁ = – 1.75/4 = – 0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.
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