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Page No 110: - Chapter 5 Laws Of Motion class 11 ncert solutions Physics - SaraNextGen [2024]


Question 5.4:

One end of a string of length is connected to a particle of mass and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed the net force on the particle (directed towards the centre) is:

(i) T, (ii) https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4432/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_6b4dab1d.gif , (iii) https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4432/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m23b9497d.gif , (iv) 0

is the tension in the string. [Choose the correct alternative].

Answer:

Answer: (i)

When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e.,

T = https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4432/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m6953f025.gif

Where F is the net force acting on the particle.

Question 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 ms–1. How long does the body take to stop?

Answer:

Retarding force, F = –50 N

Mass of the body, m = 20 kg

Initial velocity of the body, u = 15 m/s

Final velocity of the body, v = 0

Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:

F = ma

–50 = 20 × a

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4433/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_2fff3e22.gif

Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:

v = u + at

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4433/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_1e0500b0.gif  = 6 s

Question 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?

Answer:

0.18 N; in the direction of motion of the body

Mass of the body, m = 3 kg

Initial speed of the body, u = 2 m/s

Final speed of the body, v = 3.5 m/s

Time, t = 25 s

Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:

v = u + at

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4434/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m32d378a6.gif

As per Newton’s second law of motion, force is given as:

F = ma

= 3 × 0.06 = 0.18 N

Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.

Question 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.

Answer:

2 m/s2, at an angle of 37° with a force of 8 N

Mass of the body, m = 5 kg

The given situation can be represented as follows:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4435/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m7e4ddf9e.jpg

The resultant of two forces is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4435/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m23c8a7f9.gif  https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4435/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_546f95d6.gif

θ is the angle made by R with the force of 8 N

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4435/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_4eb6183a.gif

The negative sign indicates that θ is in the clockwise direction with respect to the force of magnitude 8 N.

As per Newton’s second law of motion, the acceleration (a) of the body is given as:

ma

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4435/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_2c137a35.gif

Question 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.

Answer:

Initial speed of the three-wheeler, u = 36 km/h = 10 m/s

Final speed of the three-wheeler, v = 0 m/s

Time, t = 4 s

Mass of the three-wheeler, m = 400 kg

Mass of the driver, m‘ = 65 kg

Total mass of the system, M = 400 + 65 = 465 kg

Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as:

v = u + at

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4436/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m3b90f2fc.gif

The negative sign indicates that the velocity of the three-wheeler is decreasing with time.

Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as:

F = Ma

= 465 × (–2.5) = –1162.5 N

The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.

Question 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.

Answer:

Mass of the rocket, m = 20,000 kg

Initial acceleration, a = 5 m/s2

Acceleration due to gravity, g = 10 m/s2

Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation:

F – mg = ma

F = m (g + a)

= 20000 × (10 + 5)

= 20000 × 15 = 3 × 105 N

Question 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 = 0, the position of the body at that time to be = 0, and predict its position at t = –5 s, 25 s, 100 s.

Answer:

Mass of the body, m = 0.40 kg

Initial speed of the body, u = 10 m/s due north

Force acting on the body, F = –8.0 N

Acceleration produced in the body, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4438/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m7d5b77a6.gif

(i) At t = –5 s

Acceleration, a‘ = 0 and u = 10 m/s

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4438/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m7da310f1.gif

= 10 × (–5) = –50 m

(ii) At t = 25 s

Acceleration, a” = –20 m/s2 and u = 10 m/s

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4438/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m33e5dd6d.gif

(iii) At t = 100 s

For

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4438/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_34ee160d.gif

a = –20 m/s2

u = 10 m/s

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4438/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m2aafd369.gif

= –8700 m

For

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4438/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m53ed4b61.gif

As per the first equation of motion, for t = 30 s, final velocity is given as:

v = u + at

= 10 + (–20) × 30 = –590 m/s

Velocity of the body after 30 s = –590 m/s

For motion between 30 s to 100 s, i.e., in 70 s:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4438/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m19b5e959.gif

= –590 × 70 = –41300 m

∴Total distance, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4438/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_70d708dd.gif = –50000 m

Question 5.11:

A truck starts from rest and accelerates uniformly at 2.0 m s–2. At = 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 = 11 s? (Neglect air resistance.)

Answer:

Answer: (a) 22.36 m/s, at an angle of 26.57° with the motion of the truck

(b) 10 m/s2

(a) Initial velocity of the truck, u = 0

Acceleration, a = 2 m/s2

Time, t = 10 s

As per the first equation of motion, final velocity is given as:

v = u + at

= 0 + 2 × 10 = 20 m/s

The final velocity of the truck and hence, of the stone is 20 m/s.

At t = 11 s, the horizontal component (vx) of velocity, in the absence of air resistance, remains unchanged, i.e.,

vx = 20 m/s

The vertical component (vy) of velocity of the stone is given by the first equation of motion as:

vy = u + ayδt

Where, δt = 11 – 10 = 1 s and ay = g = 10 m/s2

vy = 0 + 10 × 1 = 10 m/s

The resultant velocity (v) of the stone is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4439/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m6123f150.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4439/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m3a0887fd.gif

Let θ be the angle made by the resultant velocity with the horizontal component of velocity, vx

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4439/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_1a230638.gif

= 26.57°

(b) When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is 10 m/s2 and it acts vertically downward.

Question 5.12:

A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s–1. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.

Answer:

Answer: (a) Vertically downward

(b) Parabolic path

(a) At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.

(b)At the mean position, the velocity of the bob is 1 m/s. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.

Question 5.13:

A man of mass 70 kg stands on a weighing scale in a lift which is moving

(a) upwards with a uniform speed of 10 m s–1,

(b) downwards with a uniform acceleration of 5 m s–2,

(c) upwards with a uniform acceleration of 5 m s–2.

What would be the readings on the scale in each case?

(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Answer:

(a) Mass of the man, m = 70 kg

Acceleration, a = 0

Using Newton’s second law of motion, we can write the equation of motion as:

R – mg = ma

Where, ma is the net force acting on the man.

As the lift is moving at a uniform speed, acceleration a = 0

R = mg

= 70 × 10 = 700 N

∴Reading on the weighing scale = https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4441/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m3ff2162f.gif

(b) Mass of the man, m = 70 kg

Acceleration, a = 5 m/s2 downward

Using Newton’s second law of motion, we can write the equation of motion as:

R + mg = ma

R = m(g – a)

= 70 (10 – 5) = 70 × 5

= 350 N

∴Reading on the weighing scale = https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4441/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_6b90b350.gif

(c) Mass of the man, m = 70 kg

Acceleration, a = 5 m/s2 upward

Using Newton’s second law of motion, we can write the equation of motion as:

R – mg = ma

R = m(g + a)

= 70 (10 + 5) = 70 × 15

= 1050 N

∴Reading on the weighing scale = https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4441/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_7d5193d4.gif

(d) When the lift moves freely under gravity, acceleration a = g

Using Newton’s second law of motion, we can write the equation of motion as:

R + mg = ma

R = m(g – a)

m(g – g) = 0

∴Reading on the weighing scale = https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4441/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m5ffc7e5b.gif

The man will be in a state of weightlessness.

Question 5.14:

Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s< t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only).

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4442/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_cfc4af0.jpg

Figure 5.16

Answer:

(a) For t < 0

It can be observed from the given graph that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero.

For t > 4 s

It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of 3 m from the origin. Hence, no force is acting on the particle.

For 0 < t < 4

It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero.

(b) At t = 0

Impulse = Change in momentum

mv – mu

Mass of the particle, m = 4 kg

Initial velocity of the particle, u = 0

Final velocity of the particle, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4442/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_1d26c8bd.gif

∴Impulse https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4442/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m55f27916.gif

At t = 4 s

Initial velocity of the particle, https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4442/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m5a09aff0.gif

Final velocity of the particle, v = 0

∴ Impulse https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4442/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m261c93af.gif

Question 5.15:

Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

Answer:

Horizontal force, F = 600 N

Mass of body A, m1 = 10 kg

Mass of body B, m2 = 20 kg

Total mass of the system, m = m1 + m2 = 30 kg

Using Newton’s second law of motion, the acceleration (a) produced in the system can be calculated as:

F = ma

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4443/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_f4a6994.gif

When force F is applied on body A:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4443/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_330b723b.jpg

The equation of motion can be written as:

F – T m1a

T = F – m1a

= 600 – 10 × 20 = 400 N … (i)

When force F is applied on body B:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4443/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_161a97de.jpg

The equation of motion can be written as:

F – T = m2a

T = F – m2a

T = 600 – 20 × 20 = 200 N … (ii)

Also Read : Page-No-111:-Chapter-5-Laws-Of-Motion-class-11-ncert-solutions-Physics

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