Chapter 4: Laws of Motion Application Questions | physics Class 11 CBSE

ByteNotes

Chapter 4: Laws of Motion Application Questions | physics Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A rocket of 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. (Take g = 10 m s⁻²)

The forces on the rocket are: thrust force F upward (provided by exhaust gases) and weight mg downward; applying Newton's Second Law: F − mg = ma.
Substituting values: F = m(g + a) = 20,000 × (10 + 5) = 20,000 × 15 = 3,00,000 N = 3 × 10⁵ N.
The thrust must exceed the weight (mg = 2 × 10⁵ N) to produce net upward acceleration; the excess force (1 × 10⁵ N) provides the 5 m s⁻² upward acceleration.
This illustrates Newton's Third Law: the rocket expels exhaust gases backward with a large force; the reaction force (thrust) on the rocket acts forward (upward), propelling it.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A bullet of mass 0.04 kg moving at 90 m s⁻¹ enters a heavy wooden block and stops after travelling 60 cm. Find the average resistive force exerted by the block on the bullet.

Using v² = u² + 2as with v = 0, u = 90 m s⁻¹, s = 0.60 m: 0 = 90² + 2a(0.60), giving a = −90²/(1.20) = −6750 m s⁻².
The negative sign indicates retardation; by Newton's Second Law, the resistive force is F = ma = 0.04 × 6750 = 270 N in the direction opposing the bullet's motion.
This is an average resistive force; in reality the force may vary along the path as the bullet decelerates through the block's material.
By Newton's Third Law, the bullet exerts an equal 270 N forward force on the block, which is what causes damage to the block material.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A batsman hits back a ball straight towards the bowler without changing its speed of 12 m s⁻¹. If the ball's mass is 0.15 kg, determine the impulse imparted to the ball.

The ball reverses direction; taking the direction from batsman to bowler as positive, initial velocity = −12 m s⁻¹ and final velocity = +12 m s⁻¹.
Change in momentum = m(vf − vi) = 0.15 × (12 − (−12)) = 0.15 × 24 = 3.6 N·s in the direction from batsman to bowler.
Impulse = change in momentum = 3.6 N·s; this is the impulse imparted to the ball by the bat, directed from batsman toward bowler.
This example shows the utility of impulse: the contact time between bat and ball and the actual force during impact are difficult to measure separately, but the impulse (change in momentum) is straightforward to calculate.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

The driver of a three-wheeler (mass 400 kg) with a driver (mass 65 kg) moving at 36 km/h applies brakes and stops in 4.0 s to avoid a child. What average retarding force acts on the vehicle?

Total mass = 400 + 65 = 465 kg; initial speed u = 36 km/h = 10 m s⁻¹; final speed v = 0; time = 4.0 s.
Retardation a = (v − u)/t = (0 − 10)/4 = −2.5 m s⁻².
By Newton's Second Law, the average retarding force = m × |a| = 465 × 2.5 = 1162.5 N ≈ 1163 N directed opposite to the direction of motion.
This braking force is provided by friction between the tyres and the road; a larger friction coefficient or a longer braking distance would reduce the required force.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A cyclist is speeding at 18 km/h on a level road and takes a circular turn of radius 3 m. The coefficient of static friction between tyres and road is 0.1. Will the cyclist slip?

Convert speed: v = 18 km/h = 5 m s⁻¹; the condition for not slipping is v² ≤ μsRg.
Calculate μsRg = 0.1 × 3 × 9.8 = 2.94 m² s⁻²; calculate v² = 5² = 25 m² s⁻².
Since v² = 25 >> μsRg = 2.94, the condition is violated; the available friction is far less than the required centripetal force, so the cyclist will slip.
To avoid slipping, the cyclist should reduce speed to at most vmax = √(μsRg) = √2.94 ≈ 1.71 m s⁻¹ ≈ 6.2 km/h, or take a wider turn with larger R.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

A box of mass 50 kg lies on the floor of a train accelerating at 1.5 m s⁻². What frictional force must the floor exert on the box to keep it stationary relative to the train, and what is the minimum required coefficient of static friction?

For the box to accelerate with the train at 1.5 m s⁻², a net horizontal force must act on it; the only horizontal force available is static friction from the floor.
Required frictional force = ma = 50 × 1.5 = 75 N; this static friction acts forward on the box (in the direction of train's acceleration).
Normal force N = mg = 50 × 10 = 500 N; minimum coefficient of static friction μs ≥ f/N = 75/500 = 0.15.
If μs < 0.15, the maximum available static friction is insufficient and the box will slide backward relative to the train; this situation demonstrates that friction is the sole provider of horizontal acceleration to the box.

ByteNotes

Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

A stone of mass 0.25 kg tied to a string is whirled in a horizontal circle of radius 1.5 m at 40 rev/min. Find the tension in the string. What maximum speed can the stone have if the string withstands at most 200 N?

Angular speed ω = 40 rev/min = 40 × 2π/60 = 4π/3 rad s⁻¹; linear speed v = ωr = (4π/3) × 1.5 = 2π m s⁻¹.
Tension T = mv²/r = 0.25 × (2π)²/1.5 = 0.25 × 4π²/1.5 = π²/1.5 ≈ 6.57 N.
For maximum speed with Tmax = 200 N: Tmax = mv²max/r, so v²max = Tmax × r/m = 200 × 1.5/0.25 = 1200 m² s⁻², giving vmax = √1200 ≈ 34.6 m s⁻¹.
The tension in the string provides the centripetal force; increasing speed rapidly increases the required tension (T ∝ v²), so the string breaks when the required centripetal force exceeds its tensile limit.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

A man of mass 70 kg stands on a weighing scale in a lift. Calculate the reading on the scale when the lift (a) moves upward with uniform speed of 10 m s⁻¹, (b) accelerates upward at 5 m s⁻², (c) moves freely under gravity. (g = 10 m s⁻²)

(a) Uniform speed means zero acceleration; by Newton's Second Law, N − mg = 0, so N = mg = 700 N; scale reads 70 kg (same as the man's actual weight).
(b) Upward acceleration a = 5 m s⁻²: N − mg = ma, so N = m(g + a) = 70 × 15 = 1050 N; scale reads 105 kg (apparent weight increased—man feels heavier).
(c) Free fall: a = g = 10 m s⁻² downward; N − mg = −mg (net downward), so N = 0; scale reads zero—complete weightlessness, as both man and scale fall together.
These three cases show that the 'weight' registered on a scale is not the gravitational force but the normal reaction N, which depends on acceleration; only at rest or uniform velocity does N = mg.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

A shell of mass 0.020 kg is fired from a gun of mass 100 kg with muzzle speed 80 m s⁻¹. Find the recoil speed of the gun using conservation of momentum.

Initially both gun and shell are at rest; total initial momentum = 0.
By conservation of momentum (gun + shell system is isolated during firing): momentum of shell + momentum of gun = 0.
Let recoil speed of gun = v; then: mshell × 80 + mgun × (−v) = 0; 0.020 × 80 = 100 × v; v = 1.6/100 = 0.016 m s⁻¹.
The recoil velocity is very small because the gun is much more massive; this illustrates that in an isolated system, momentum is transferred between bodies, and a massive body acquires a much smaller velocity than a light body for the same impulse.

ByteNotes

Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 10

Applied Concept

A constant force acts on a body of mass 3.0 kg and changes its speed from 2.0 m s⁻¹ to 3.5 m s⁻¹ in 25 s, direction remaining unchanged. Find the magnitude and direction of the force.

Acceleration a = (v − u)/t = (3.5 − 2.0)/25 = 1.5/25 = 0.06 m s⁻².
By Newton's Second Law: F = ma = 3.0 × 0.06 = 0.18 N.
Since the speed increases and direction is unchanged, the force is in the direction of motion; F = 0.18 N in the direction of the body's motion.
This small force applied over 25 s produces a significant change in speed, demonstrating the cumulative effect of even a small constant force over time—the impulse = 0.18 × 25 = 4.5 N·s equals the change in momentum = 3.0 × 1.5 = 4.5 kg m s⁻¹.

ByteNotes

Premium Access

Unlock all Application Questions

You are viewing 10 free application questions from this chapter. Upgrade to Premium to unlock the remaining 13 questions.

Upgrade to Premium13 questions locked

Chapter sections

Chapter 4: Laws of Motion | Application Questions

A rocket of 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. (Take g = 10 m s⁻²)

A bullet of mass 0.04 kg moving at 90 m s⁻¹ enters a heavy wooden block and stops after travelling 60 cm. Find the average resistive force exerted by the block on the bullet.

A batsman hits back a ball straight towards the bowler without changing its speed of 12 m s⁻¹. If the ball's mass is 0.15 kg, determine the impulse imparted to the ball.

The driver of a three-wheeler (mass 400 kg) with a driver (mass 65 kg) moving at 36 km/h applies brakes and stops in 4.0 s to avoid a child. What average retarding force acts on the vehicle?

A cyclist is speeding at 18 km/h on a level road and takes a circular turn of radius 3 m. The coefficient of static friction between tyres and road is 0.1. Will the cyclist slip?

A box of mass 50 kg lies on the floor of a train accelerating at 1.5 m s⁻². What frictional force must the floor exert on the box to keep it stationary relative to the train, and what is the minimum required coefficient of static friction?

A stone of mass 0.25 kg tied to a string is whirled in a horizontal circle of radius 1.5 m at 40 rev/min. Find the tension in the string. What maximum speed can the stone have if the string withstands at most 200 N?

A man of mass 70 kg stands on a weighing scale in a lift. Calculate the reading on the scale when the lift (a) moves upward with uniform speed of 10 m s⁻¹, (b) accelerates upward at 5 m s⁻², (c) moves freely under gravity. (g = 10 m s⁻²)

A shell of mass 0.020 kg is fired from a gun of mass 100 kg with muzzle speed 80 m s⁻¹. Find the recoil speed of the gun using conservation of momentum.

A constant force acts on a body of mass 3.0 kg and changes its speed from 2.0 m s⁻¹ to 3.5 m s⁻¹ in 25 s, direction remaining unchanged. Find the magnitude and direction of the force.

Unlock all Application Questions