Laws of Motion — Maharashtra (SSC) Class 9 Science Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 9 Science solutions for "Laws of Motion" — important questions with detailed answers, download PDF for board exam preparation.

Laws of Motion — Maharashtra (SSC) Class 9 Science Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 9 Science solutions for "Laws of Motion" — important questions with detailed answers, download PDF for board exam preparation.

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TL;DR: Free step-by-step Maharashtra (SSC) Class 9 Science solutions for "Laws of Motion" — important questions with detailed answers, download PDF for board…

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Laws of Motion — Textbook Questions Answered (9)

Q1: State Newton's First Law of Motion, and explain why a passenger in a moving bus lurches forward when the bus suddenly stops.

Newton's First Law: an object at rest stays at rest, and an object in motion continues moving at constant velocity in a straight line, unless acted upon by an external unbalanced force. This tendency of an object to resist any change in its state of motion is called inertia. Why the passenger lurches forward: Before the bus brakes, both the bus and the passenger's body are moving forward together at the same speed. When the bus suddenly stops, an external force (friction from the brakes) acts on the BUS, stopping it quickly. However, no such external force directly acts on the passenger's body at that instant — by the First Law, the passenger's body simply continues moving forward at its previous speed, due to its own inertia, until the seatbelt, seat, or the passenger's own muscles apply a force to stop it too. This is exactly why seatbelts exist: they provide the necessary external force to stop the passenger's body safely, rather than relying on a sudden collision with the dashboard or windscreen to do it.

Q2: A force of 10 N is applied to a 2 kg mass. Calculate the acceleration produced.

By Newton's Second Law: F = ma, where F is the net force, m is mass, and a is the resulting acceleration. Given: F = 10 N, m = 2 kg Rearranging for acceleration: a = F/m = 10/2 = 5 m/s² The acceleration produced is 5 m/s². This means the object's velocity increases by 5 metres per second, every second, for as long as this 10 N force continues to act.

Q3: Define momentum and state the Law of Conservation of Momentum.

Momentum (p) is a measure of the quantity of motion an object has, defined as the product of its mass and velocity: p = mv (measured in kg·m/s) A heavier object moving at the same speed as a lighter one has more momentum, and a faster object has more momentum than a slower one of the same mass — momentum captures both factors together, which is why it takes noticeably more effort to stop a heavy truck than a bicycle moving at the same speed. Law of Conservation of Momentum: in a closed system where no external force acts, the total momentum before an event (such as a collision or explosion) equals the total momentum after it. This law follows directly from Newton's Third Law: whenever two objects interact, the force one exerts on the other is exactly equal and opposite to the force the second exerts back, for exactly the same duration of contact. Equal and opposite forces acting for equal times produce equal and opposite changes in momentum, which is why the total momentum of the whole system stays unchanged even though it may be redistributed between the individual objects.

Q4: Two balls of mass 3 kg and 2 kg move at 4 m/s and 6 m/s respectively in opposite directions. Calculate the total momentum of the system.

Since the two balls move in opposite directions, one direction must be taken as positive and the other as negative. Given: m₁ = 3 kg, v₁ = +4 m/s (taking this direction as positive); m₂ = 2 kg, v₂ = −6 m/s (opposite direction, so negative) Step 1: Find each ball's momentum p₁ = m₁v₁ = 3 × 4 = 12 kg·m/s p₂ = m₂v₂ = 2 × (−6) = −12 kg·m/s Step 2: Add them (momentum is a vector quantity, so direction/sign matters) Total momentum = p₁ + p₂ = 12 + (−12) = 0 kg·m/s The total momentum of the system is zero — the two balls' momenta exactly cancel out, even though neither ball individually has zero momentum. This is a common exam trap: students sometimes add magnitudes without accounting for direction and get 20 kg·m/s instead, which ignores that momentum is a vector.

Q5: State Newton's Third Law of Motion and explain what an action-reaction pair is, using a wall-pushing example.

Newton's Third Law: for every action, there is an equal and opposite reaction — whenever one object exerts a force on a second object, the second object simultaneously exerts an equal force in the opposite direction back on the first. Example — pushing a wall: When you push against a wall, you exert a force ON the wall. At the exact same instant, the wall pushes back on YOU with a force equal in magnitude but opposite in direction — this is why your hand feels resistance and you can even push yourself backward off a wall. Key features of an action-reaction pair: 1. The two forces act on TWO DIFFERENT bodies (one on the wall, one on you) — this is often misunderstood; they never act on the same object, so they never cancel each other out for either body individually. 2. They are equal in magnitude. 3. They act in exactly opposite directions. 4. They occur simultaneously — neither force exists before or after the other. Because the two forces act on different objects, and each object may have very different mass, they can produce very different accelerations even though the forces themselves are equal — this is exactly why a swimmer pushing backward against the water (the action) is propelled forward by the water's reaction force, even though the swimmer's mass is far smaller than the mass of water they push against.

Q6: A 1500 kg car accelerates from rest to 20 m/s in 5 seconds. Calculate the force applied by the engine.

Step 1: Find the acceleration using a = (v − u)/t u = 0 (starts from rest), v = 20 m/s, t = 5 s a = (20 − 0)/5 = 4 m/s² Step 2: Apply Newton's Second Law, F = ma F = 1500 × 4 = 6,000 N The force applied by the car's engine is 6,000 newtons (assuming this is the net force after accounting for friction and air resistance — in reality the engine must generate somewhat more force than this to also overcome those opposing resistive forces).

Q7: Explain inertia and its relationship to mass. Why is a loaded truck harder to accelerate than an empty car, given the same applied force?

Inertia is the natural tendency of any object to resist a change in its current state of motion — whether that means resisting being set into motion from rest, resisting being slowed down or stopped, or resisting a change of direction. The relationship to mass: inertia is directly proportional to mass — the greater an object's mass, the greater its inertia, and therefore the more it resists any change in motion. This is essentially what Newton's Second Law (a = F/m) expresses mathematically: for the SAME applied force F, an object with LARGER mass m will experience a SMALLER acceleration a. Why the truck is harder to accelerate: Given the exact same applied force, a heavily loaded truck (large mass, large inertia) will accelerate far less than an empty, lighter car (small mass, small inertia) — the truck's greater inertia means the same push simply produces a much smaller change in its motion. This is precisely why heavy vehicles require much more powerful engines and much longer braking distances than lighter ones: both speeding up and slowing down require overcoming that larger inertia.

Q8: A force of 50 N acts on a 0.5 kg ball, initially at rest, for 0.2 seconds. Find the final velocity of the ball using the impulse-momentum relationship.

Impulse is defined as the product of force and the time for which it acts, and by Newton's Second Law, impulse equals the resulting change in momentum: Impulse = F × t = change in momentum = m(v − u) Given: F = 50 N, t = 0.2 s, m = 0.5 kg, u = 0 (initially at rest) Step 1: Calculate the impulse Impulse = F × t = 50 × 0.2 = 10 kg·m/s Step 2: Set this equal to the change in momentum and solve for v 10 = 0.5 × (v − 0) v = 10/0.5 = 20 m/s The ball's final velocity is 20 m/s.

Q9: A gun of mass 4 kg fires a bullet of mass 0.02 kg at a velocity of 400 m/s. Find the recoil velocity of the gun, using the law of conservation of momentum.

Before firing, both the gun and the bullet are at rest, so the total momentum of the system is zero. By the law of conservation of momentum, the total momentum must still be zero immediately AFTER firing too, since no external force has acted on the gun-bullet system. Let the recoil velocity of the gun be v (in the opposite direction to the bullet). Momentum of bullet + Momentum of gun = 0 (0.02 × 400) + (4 × v) = 0 8 + 4v = 0 v = −8/4 = −2 m/s The negative sign shows the gun moves in the OPPOSITE direction to the bullet, with a recoil speed of 2 m/s. This is the exact physical reason a gun "kicks back" into the shooter's shoulder when fired — the gun's much larger mass compared to the bullet means it only needs a much smaller recoil speed to carry the same magnitude of momentum in the opposite direction.

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