Model Test – Solved ഉത്തരങ്ങൾ

Momentum (മൊമെന്റം) p = mv, so its unit is kg × m/s = kg m/s (the textbook writes kgm/s).

Force, momentum and velocity have a direction (സദിശ അളവ്, vector). Speed (വേഗം) has only a size, no direction.

With no friction (ഘർഷണം) and no other unbalanced external force, nothing changes its motion – it continues in uniform motion in a straight line (നേർരേഖയിലുള്ള സമചലനം). This idea led to Newton’s first law.

The bus stops, but the passengers’ bodies tend to keep moving forward – inertia of motion (ചലനജഡത്വം).

Inertia (ജഡത്വം) increases with mass (മാസ്) – the sand-filled bottle did not fall over, the empty one did. It does not depend on speed.

Impulse (ബലത്തിന്റെ ആവേഗം) I = F × t, so its unit is newton × second = N s (which equals kg m/s).

Take east as positive: 30 + (−20) = +10 N, i.e. 10 N towards the east – the side of the bigger force. The forces are unbalanced (അസന്തുലിതബലങ്ങൾ).

Action and reaction (ബലവും പ്രതിബലവും) always act on two different bodies, at the same instant. That is why they never cancel each other.

Newton’s first law (ന്യൂട്ടന്റെ ഒന്നാം ചലനനിയമം): every object continues in its state of rest or of uniform motion in a straight line unless an unbalanced external force acts on it.

It defines force (ബലം) – that which changes or tends to change the state of rest or uniform motion of an object – and inertia (ജഡത്വം) – the tendency of an object to continue in its state of rest or motion.

Given: m = 1500 kg; v = 72 km/h = 72 × 5/18 m/s = 20 m/s

Formula: p = mv

Substitution: p = 1500 × 20

Answer: p = 30,000 kg m/s (in the direction of motion of the car).

Given: F = 60 N, m = 12 kg, u = 0, t = 3 s

Formula: (a) F = ma, so a = F/m   (b) v = u + at

Substitution: (a) a = 60/12   (b) v = 0 + 5 × 3

Answer: (a) a = 5 m/s²   (b) v = 15 m/s

The ball’s momentum has to be brought to zero; this change of momentum (മൊമെന്റവ്യത്യാസം) is fixed. By the impulse–momentum principle (ആവേഗമൊമെന്റതത്വം) Ft = mv − mu, so F = (mv − mu)/t.

Moving the hands back increases the time t taken to stop the ball, so the force on the hands becomes smaller and the hands are not hurt.

(a) The branch suddenly moves, but the mangoes tend to remain at rest because of inertia of rest (നിശ്ചലജഡത്വം); their stalks break and they fall.

(b) To jump forward, the person pushes the boat backwards (action); the boat pushes the person forwards (reaction). By Newton’s third law the boat, which is free to move on water, moves backwards.

(a) No. The two pulls are equal and opposite, so the resultant is zero – the forces are balanced (സന്തുലിതബലങ്ങൾ).

(b) The car moves towards the left. The left pull is now bigger, so the resultant is not zero – an unbalanced force (അസന്തുലിതബലം) acts towards the left and changes the car’s state of rest; it moves in the direction of the resultant. (In the real activity the extra pull must also overcome a little friction.)

(c) External forces (ബാഹ്യബലങ്ങൾ) – they are applied on the car from outside, through the strings.

Given: m = 500 g = 0.5 kg, u = 0, v = 20 m/s, t = 0.05 s

Formula: (a) Δp = mv − mu   (b) F = (mv − mu)/t   (c) I = Ft

Substitution: (a) Δp = 0.5 × 20 − 0.5 × 0   (b) F = 10/0.05   (c) I = 200 × 0.05

Answer: (a) 10 kg m/s   (b) F = 200 N   (c) I = 10 N s – equal to the change in momentum, as the impulse–momentum principle says.

(a) Both read 50 N, in opposite directions: A pulls B with 50 N and B pulls A back with 50 N.

(b) Newton’s third law (ന്യൂട്ടന്റെ മൂന്നാം ചലനനിയമം): for every action there is an equal and opposite reaction.

(c) Forces can cancel only if they act on the same body. The action acts on B and the reaction on A – different bodies – so they cannot cancel.

Read the graph: v goes 0 → 12 m/s in 0–4 s, stays 12 m/s from 4–8 s, then 12 → 0 m/s in 8–10 s. Acceleration a = (v − u)/t, then F = ma.

Given: m = 2 kg

Formula: a = (v − u)/t,   F = ma

Substitution: (a) a = (12 − 0)/4 = 3 m/s², F = 2 × 3   (b) a = (12 − 12)/4 = 0, F = 2 × 0   (c) a = (0 − 12)/2 = −6 m/s², F = 2 × (−6)

Answer: (a) 6 N   (b) 0 N (uniform velocity – balanced forces)   (c) −12 N

The negative sign shows that in (c) the force acts opposite to the direction of motion – it is a retarding force (മന്ദീകരണം, retardation 6 m/s²).

Given: m = 10,000 kg; u = 54 km/h = 54 × 5/18 = 15 m/s; v = 0; t = 5 s

Formula: (a) p = mu   (b) F = (mv − mu)/t   (c) a = (v − u)/t,   v² = u² + 2as

Substitution: (a) p = 10,000 × 15   (b) F = 10,000 × (0 − 15)/5 = −150,000/5   (c) a = (0 − 15)/5 = −3 m/s²; 0 = 15² + 2 × (−3) × s, so s = 225/6

Answer: (a) 150,000 kg m/s   (b) F = −30,000 N – the minus sign shows the force acts opposite to the motion (a braking force).   (c) retardation 3 m/s², stopping distance s = 37.5 m

(d) u = 108 km/h = 30 m/s, a = −3 m/s²: s = u²/2a = 30²/(2 × 3) = 900/6 = 150 m – four times as far. Doubling the speed makes the stopping distance four times (s ∝ u²): overspeeding is dangerous.

(e) The bus stops, but the passengers’ bodies tend to continue moving forward – inertia of motion (ചലനജഡത്വം).

(a) By Newton’s third law, B pushes A with an equal force of 120 N in the opposite direction, at the same time.

Given: m_A = 40 kg, m_B = 60 kg, F = 120 N on each (opposite directions), u = 0, t = 0.5 s

Formula: (b) a = F/m   (c) v = u + at   (d) p = mv

Substitution: (b) a_A = 120/40, a_B = 120/60   (c) v_A = 0 + 3 × 0.5, v_B = 0 + 2 × 0.5   (d) p_A = 40 × 1.5, p_B = 60 × 1

Answer: (b) a_A = 3 m/s², a_B = 2 m/s²   (c) v_A = 1.5 m/s, v_B = 1 m/s, in opposite directions   (d) 60 kg m/s each, in opposite directions

(d) Taking B’s direction as positive: total momentum = 60 + (−60) = 0 – the same as before the push (both were at rest).

(e) Newton’s third law: action (on B) and reaction (on A) act on different bodies, so each student gets a force and moves. The forces are equal, but A has the smaller mass, so by F = ma (second law) A gets the larger acceleration and moves faster.