Laws of Motion

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New answer posted•a year ago

Laws of Motion Class 11th

5.11 A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)

16 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

The acceleration of the truck, a = 2 m/s², t = 10 s, Initial velocity, u = 0, from the equation v = u + at, we get v = 2 $\times 10$ = 20 m/s

At time t = 11 s, the horizontal component of the velocity Vx remains unchanged due to no air resistance. Hence Vx = 20 m/s

The vertical component of the velocity Vy

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New answer posted•a year ago

Laws of Motion Class 11th

5.10 A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.

154 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Given, the mass of the body = 0.4 kg, Initial speed = 10m/s, Initial force, Retarding force, F = - 8 N

(a) At t = -5s, Due to constant speed, the acceleration of the body = 0

From the relation, s = ut + (1/2)at², we get

s = 10 $\times (- 5)$ + 0 $\times$ ${(\frac{1}{2}) (- 5)}^{2}$ = -50 m

At t = 25s, The acceleration o

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New answer posted•a year ago

Laws of Motion Class 11th

5.9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s^-2. Calculate the initial thrust (force) of the blast.

21 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

The mass of the rocket, m = 20000 kg

When the rocket is fired, gravitational acceleration tries to pull it down. Hence the effective acceleration on the rocket = rocket acceleration + gravitational acceleration

The acceleration, a = 5 m/s², gravitational acceleration =10 m/s²

Total acceleration

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New answer posted•a year ago

Laws of Motion Class 11th

5.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.

30 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

The initial speed of 3 wheeler, u = 36 km/h = 10 m/s

The final velocity of the 3 wheeler, v = 0

The time t = 4 s

From the relation, v = u-at, we get, a = u/t = 10/4 m/s²= 2.5 m/s²

The total mass acting on the 3 wheeler, m = mass of the 3 wheeler + mass of the driver = 400 +65 kg = 465 kg

The average ret

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New answer posted•a year ago

Laws of Motion Class 11th

5.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

35 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Given, the mass m = 5 kg

The resultant force F of the two forces 6N and 8N is given by,

F = $\sqrt{8^{2} + 6^{2}}$ = 10 N

Acceleration is given by F = ma, a = F/m = 10/5 m/s² = 2 m/s²

$\tan ? θ$ = 6/8, $θ$ = 36.86 $°$

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New answer posted•a year ago

Laws of Motion Class 11th

5.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s^-1 to 3.5 m s^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

11 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Initial speed, u = 2.0 m/s, the final speed v = 3.5 m/s, the time t = 25 s

From the relation v = u + at, we get acceleration a = (v-u)/t = (3.5 – 2.0 )/ 25s = 0.06 m/s²

The force F = ma, F = 3 $*$ 0.06 N = 0.18 N

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New answer posted•a year ago

Laws of Motion Class 11th

5.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s-1. How long does the body take to stop?

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

The retarding force, F = 50N, the mass m = 20 kg

From the equation F = ma, we get acceleration a = 50 / 20 m/s² = 2.5 m/s²

The initial speed, u = 15 m/s, the final speed v =0, from the relation v = u-at, we get

T = u/a = 15/2.5 s = 6s

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New answer posted•a year ago

Laws of Motion Class 11th

5.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,

(a) Just after it is dropped from the window of a stationary train

(b) Just after it is dropped from the window of a train running at a constant velocity of 36 km/h

(c) Just after it is dropped from the window of a train accelerating with 1 m s^-2

(d) Lying on the floor of a train which is accelerating with 1 m s^-2, the stone being at rest relative to the train. (Neglect air resistance throughout.)

83 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

(a) Mass of the stone, m = 0.1 kg, Acceleration due to gravity = 10 m/s²

Net force = 0.1 $*$ 10 N = 1.0 N. The direction of force is downwards

(b) Since the train is moving at a constant velocity, the acceleration imparted by the train is zero. Only acceleration acting on the stone is gravit

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New answer posted•a year ago

Laws of Motion Class 11th

5.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,

(a) During its upward motion

(b) During its downward motion

(c) At the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?

(Ignore air resistance.)

272 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

(a) During the upward motion f the pebble, the acceleration due to gravity acts downwards. The magnitude of the force is

F = MA = 0.05 kg $*$ 10 m/s² = 0.5 N and the direction of force is downwards.

(b) During its downward motion – the force is 0.5 N and the direction is downwards.

(c) When the pe

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New answer posted•a year ago

Laws of Motion Class 11th

5.1 Give the magnitude and direction of the net force acting on

(a) A drop of rain falling down with a constant speed

(b) A cork of mass 10 g floating on water

(c) A kite skillfully held stationary in the sky

(d) A car moving with a constant velocity of 30 km/h on a rough road

(e) A high-speed electron in space far from all material objects, and free of electric and magnetic fields

274 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

(a) Since the raindrop is falling at a constant speed, the acceleration is zero. When the acceleration is 0, the force acting on the drop will also become zero ( since F = ma)

(b) Since the cork is floating on water, the weight of the cork is balanced by upward force of water. So the net force on t

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