Force and Laws of Motion

Given/Concept: Net external unbalanced force = 0. By Newton's First Law of Motion, an object continues in its state of rest or of uniform motion in a straight line unless acted upon by an unbalanced external force.

Answer: Yes, it is absolutely possible for the object to be travelling with a non-zero velocity even when the net external unbalanced force is zero.

Conditions:
- The object must be moving with a constant (uniform) velocity — i.e., no change in speed.
- The direction of motion must remain unchanged (straight line).
- There should be no acceleration (since $F = ma$, if $F = 0$ then $a = 0$).

For example, a hockey puck sliding on a frictionless ice surface moves with constant velocity because no net unbalanced force acts on it.

Concept Used: Newton's First Law of Motion (Law of Inertia).

Explanation:
- When a carpet is beaten with a stick, the carpet is suddenly set into motion.
- The dust particles on the carpet, due to their inertia of rest, tend to remain in their original position (at rest).
- Since the carpet moves forward but the dust particles stay behind (due to inertia), the dust gets separated from the carpet and falls off.

This is a direct application of the property of inertia — every object resists a change in its state of rest or motion.

Concept Used: Newton's First Law of Motion (Inertia of motion and inertia of rest).

Explanation:
- When the bus suddenly accelerates (starts moving), the luggage on the roof tends to remain at rest due to inertia of rest, and may fall backward.
- When the bus suddenly brakes (stops), the luggage tends to continue moving forward due to inertia of motion, and may fall forward.
- When the bus takes a sharp turn, the luggage tends to continue in a straight line due to inertia and may slide off sideways.

Therefore, it is advised to tie the luggage with a rope so that it moves along with the bus and does not fall off due to inertia.

Correct Option: (c) there is a force on the ball opposing the motion.

Justification: After the ball leaves the bat, the only horizontal force acting on it is the force of friction between the ball and the ground, which opposes the motion of the ball. This unbalanced frictional force decelerates the ball and eventually brings it to rest. Options (a), (b), and (d) are incorrect because the ball's stopping has nothing to do with how hard it was hit, velocity is not proportional to force (acceleration is), and there IS an unbalanced force (friction) acting on the ball.

Given:
- Initial velocity, $u = 0$ (starts from rest)
- Distance, $s = 400$ m
- Time, $t = 20$ s
- Mass, $m = 7$ tonnes $= 7 \times 1000 = 7000$ kg

Step 1: Find acceleration using the equation of motion $s = ut + \frac{1}{2}at^2$

$$s = ut + \frac{1}{2}at^2$$
$$400 = 0 \times 20 + \frac{1}{2} \times a \times (20)^2$$
$$400 = \frac{1}{2} \times a \times 400$$
$$400 = 200a$$
$$a = \frac{400}{200} = 2 \text{ m s}^{-2}$$

Step 2: Find force using Newton's Second Law $F = ma$

$$F = ma = 7000 \times 2 = 14000 \text{ N}$$

Answer: Acceleration $= 2 \text{ m s}^{-2}$ and Force $= 14000 \text{ N}$.

Given:
- Mass of stone, $m = 1$ kg
- Initial velocity, $u = 20$ m s$^{-1}$
- Final velocity, $v = 0$ (comes to rest)
- Distance, $s = 50$ m

Step 1: Find acceleration using $v^2 = u^2 + 2as$

$$v^2 = u^2 + 2as$$
$$0 = (20)^2 + 2 \times a \times 50$$
$$0 = 400 + 100a$$
$$a = \frac{-400}{100} = -4 \text{ m s}^{-2}$$

(Negative sign indicates deceleration/retardation)

Step 2: Find force of friction using $F = ma$

$$F = ma = 1 \times (-4) = -4 \text{ N}$$

The negative sign indicates the force opposes the motion.

Answer: The force of friction between the stone and the ice is $4$ N (opposing the motion).

Given:
- Mass of engine, $m_e = 8000$ kg
- Number of wagons $= 5$, each of mass $= 2000$ kg
- Total mass of wagons $= 5 \times 2000 = 10000$ kg
- Total mass of train (engine + wagons), $m = 8000 + 10000 = 18000$ kg
- Force exerted by engine $= 40000$ N
- Friction force $= 5000$ N

(a) Net accelerating force:

$$F_{net} = F_{engine} - F_{friction}$$
$$F_{net} = 40000 - 5000 = 35000 \text{ N}$$

(b) Acceleration of the train:

Using Newton's Second Law: $F = ma$
$$a = \frac{F_{net}}{m} = \frac{35000}{18000}$$
$$a \approx 1.944 \approx 1.94 \text{ m s}^{-2}$$

Answer: (a) Net accelerating force $= 35000$ N; (b) Acceleration $\approx 1.94$ m s$^{-2}$.

Given:
- Mass of vehicle, $m = 1500$ kg
- Acceleration (retardation), $a = -1.7$ m s$^{-2}$

Using Newton's Second Law: $F = ma$

$$F = 1500 \times (-1.7)$$
$$F = -2550 \text{ N}$$

The negative sign indicates that the force acts opposite to the direction of motion (i.e., it is a braking/retarding force).

Answer: The force between the vehicle and the road must be $2550$ N (acting opposite to the direction of motion).

Correct Option: (d) $mv$

Justification: By definition, momentum ($p$) is the product of the mass of an object and its velocity:
$$p = mv$$
It is a vector quantity with the same direction as the velocity. Its SI unit is kg m s$^{-1}$. Options (a), (b), and (c) represent other physical quantities (related to kinetic energy, etc.) and are not momentum.

Given:
- Applied horizontal force $= 200$ N
- The cabinet moves at constant velocity (i.e., zero acceleration)

Concept: Since the cabinet moves at constant velocity, the net force on it is zero (Newton's First Law).

$$F_{net} = F_{applied} - F_{friction} = 0$$
$$F_{friction} = F_{applied} = 200 \text{ N}$$

Answer: The friction force exerted on the cabinet is $200$ N, acting in the direction opposite to the motion.

Comment on the student's logic: The student's logic is incorrect. The action and reaction forces in Newton's Third Law always act on two different bodies — they never act on the same body. Therefore, they cannot cancel each other.

Correct Explanation:
- When we push the truck (action force on the truck), the truck pushes back on us (reaction force on us). These two forces act on different bodies (truck and person), so they do NOT cancel each other.
- The truck does not move because the force we apply on the truck is not large enough to overcome the static friction between the truck's tyres and the road.
- The net force on the truck = Applied force − Static friction force. Since static friction is large enough to balance our push, the net force on the truck is zero, and it does not accelerate.

Conclusion: The truck remains stationary not because action and reaction cancel, but because the applied force is insufficient to overcome static friction acting on the truck.

Given:
- Mass of hockey ball, $m = 200$ g $= 0.2$ kg
- Initial velocity, $u = 10$ m s$^{-1}$ (let this be positive direction)
- Final velocity, $v = -5$ m s$^{-1}$ (returns along original path, so negative direction)

Initial momentum:
$$p_i = mu = 0.2 \times 10 = 2 \text{ kg m s}^{-1}$$

Final momentum:
$$p_f = mv = 0.2 \times (-5) = -1 \text{ kg m s}^{-1}$$

Change in momentum:
$$\Delta p = p_f - p_i = -1 - 2 = -3 \text{ kg m s}^{-1}$$

Magnitude of change in momentum:
$$|\Delta p| = 3 \text{ kg m s}^{-1}$$

Answer: The magnitude of change of momentum of the hockey ball is $3$ kg m s$^{-1}$.

Given:
- Mass of bullet, $m = 10$ g $= 0.01$ kg
- Initial velocity, $u = 150$ m s$^{-1}$
- Final velocity, $v = 0$ (comes to rest)
- Time, $t = 0.03$ s

Step 1: Find acceleration (retardation)

$$a = \frac{v - u}{t} = \frac{0 - 150}{0.03} = \frac{-150}{0.03} = -5000 \text{ m s}^{-2}$$

Step 2: Find distance of penetration using $s = ut + \frac{1}{2}at^2$

$$s = ut + \frac{1}{2}at^2$$
$$s = 150 \times 0.03 + \frac{1}{2} \times (-5000) \times (0.03)^2$$
$$s = 4.5 + \frac{1}{2} \times (-5000) \times 0.0009$$
$$s = 4.5 - 2.25 = 2.25 \text{ m}$$

Alternatively using $v^2 = u^2 + 2as$:
$$0 = (150)^2 + 2 \times (-5000) \times s$$
$$s = \frac{22500}{10000} = 2.25 \text{ m}$$

Step 3: Find force exerted by wooden block on bullet

$$F = ma = 0.01 \times (-5000) = -50 \text{ N}$$

Magnitude of force $= 50$ N (opposing the motion of the bullet).

Answer: Distance of penetration $= 2.25$ m; Force exerted by wooden block on bullet $= 50$ N.

Given:
- Mass of object, $m_1 = 1$ kg; initial velocity, $u_1 = 10$ m s$^{-1}$
- Mass of wooden block, $m_2 = 5$ kg; initial velocity, $u_2 = 0$ (stationary)
- After collision, they stick together: combined mass $= m_1 + m_2 = 6$ kg

Total momentum just before impact:
$$p_{before} = m_1 u_1 + m_2 u_2 = (1 \times 10) + (5 \times 0) = 10 \text{ kg m s}^{-1}$$

By the Law of Conservation of Momentum:
Total momentum is conserved (no external unbalanced force).
$$p_{after} = p_{before} = 10 \text{ kg m s}^{-1}$$

Velocity of the combined object after collision:
$$p_{after} = (m_1 + m_2) \times v$$
$$10 = 6 \times v$$
$$v = \frac{10}{6} = \frac{5}{3} \approx 1.67 \text{ m s}^{-1}$$

Answer:
- Total momentum just before impact $= 10$ kg m s$^{-1}$
- Total momentum just after impact $= 10$ kg m s$^{-1}$
- Velocity of combined object $\approx 1.67$ m s$^{-1}$

Given:
- Mass, $m = 100$ kg
- Initial velocity, $u = 5$ m s$^{-1}$
- Final velocity, $v = 8$ m s$^{-1}$
- Time, $t = 6$ s

Initial momentum:
$$p_i = mu = 100 \times 5 = 500 \text{ kg m s}^{-1}$$

Final momentum:
$$p_f = mv = 100 \times 8 = 800 \text{ kg m s}^{-1}$$

Force exerted on the object (using Newton's Second Law):
$$F = \frac{p_f - p_i}{t} = \frac{800 - 500}{6} = \frac{300}{6} = 50 \text{ N}$$

Answer:
- Initial momentum $= 500$ kg m s$^{-1}$
- Final momentum $= 800$ kg m s$^{-1}$
- Force exerted on the object $= 50$ N

Analysis of each suggestion:

Kiran's suggestion (insect suffered greater change in momentum): This is incorrect.
By the Law of Conservation of Momentum, the change in momentum of the insect is equal in magnitude to the change in momentum of the motorcar. The total momentum of the system is conserved. Although the change in velocity of the insect is much larger (because its mass is very small), the change in momentum ($\Delta p = m \Delta v$) is the same in magnitude for both.

Akhtar's suggestion (motorcar exerted larger force on insect): This is also incorrect.
By Newton's Third Law of Motion, the force exerted by the motorcar on the insect is equal in magnitude to the force exerted by the insect on the motorcar. The forces are equal and opposite. The insect died because its small mass resulted in a very large acceleration (deceleration) when the same force acted on it.

Rahul's suggestion (both experienced the same force and change in momentum): This is correct.
- By Newton's Third Law, the force exerted on the insect by the car equals the force exerted on the car by the insect (equal and opposite).
- By the Law of Conservation of Momentum, the magnitude of change in momentum of the insect equals the magnitude of change in momentum of the motorcar.

Conclusion: Rahul's explanation is scientifically correct. The insect and the motorcar experience equal forces (Newton's Third Law) and equal magnitudes of change in momentum (Conservation of Momentum). The insect dies because the same force causes a much larger acceleration in the insect due to its very small mass.

Given:
- Mass of dumb-bell, $m = 10$ kg
- Height, $h = 80$ cm $= 0.80$ m
- Initial velocity, $u = 0$ (falls from rest)
- Acceleration due to gravity, $g = 10$ m s$^{-2}$

Step 1: Find velocity just before hitting the floor using $v^2 = u^2 + 2gh$

$$v^2 = 0 + 2 \times 10 \times 0.80 = 16$$
$$v = 4 \text{ m s}^{-1}$$

Step 2: Calculate momentum just before hitting the floor

$$p = mv = 10 \times 4 = 40 \text{ kg m s}^{-1}$$

The dumb-bell comes to rest after hitting the floor (final momentum = 0).

Momentum transferred to the floor:
$$\Delta p = 40 - 0 = 40 \text{ kg m s}^{-1}$$

Answer: The dumb-bell will transfer a momentum of $40$ kg m s$^{-1}$ to the floor.

Observation: The distance values are $0, 1, 8, 27, 64, 125, 216, 343$, which are $0^3, 1^3, 2^3, 3^3, 4^3, 5^3, 6^3, 7^3$.

So, $s = t^3$.

Finding velocity and acceleration:

Velocity: $v = \frac{ds}{dt} = 3t^2$ (velocity increases with time — non-uniform)

Acceleration: $a = \frac{dv}{dt} = 6t$ (acceleration increases with time)

Let us verify by calculating distances covered in successive seconds:
- $0$ to $1$ s: $1 - 0 = 1$ m
- $1$ to $2$ s: $8 - 1 = 7$ m
- $2$ to $3$ s: $27 - 8 = 19$ m
- $3$ to $4$ s: $64 - 27 = 37$ m
- $4$ to $5$ s: $125 - 64 = 61$ m
- $5$ to $6$ s: $216 - 125 = 91$ m
- $6$ to $7$ s: $343 - 216 = 127$ m

The distances covered in successive equal time intervals keep increasing by larger and larger amounts, confirming that acceleration is increasing (non-uniform acceleration).

(a) Conclusion about acceleration: The acceleration is increasing with time (non-uniform/non-constant acceleration).

(b) Inference about forces: Since the acceleration is increasing (non-uniform), the net force acting on the object is also increasing with time. The force is not constant — it increases as time progresses.

Given:
- Mass of motorcar, $m = 1200$ kg
- Two persons push at uniform velocity (acceleration $= 0$)
- Three persons push with acceleration $= 0.2$ m s$^{-2}$
- Each person exerts the same force $F$.

Step 1: Find friction force using the two-person case.

When two persons push at uniform velocity, net force $= 0$:
$$2F - f = 0 \quad \Rightarrow \quad f = 2F$$

where $f$ is the friction force.

Step 2: Apply Newton's Second Law for the three-person case.

$$3F - f = ma$$
$$3F - 2F = 1200 \times 0.2$$
$$F = 240 \text{ N}$$

Step 3: Verify friction force:
$$f = 2F = 2 \times 240 = 480 \text{ N}$$

Answer: Each person pushes the motorcar with a force of $240$ N.

Given:
- Mass of hammer, $m = 500$ g $= 0.5$ kg
- Initial velocity, $u = 50$ m s$^{-1}$
- Final velocity, $v = 0$ (hammer stops)
- Time, $t = 0.01$ s

Step 1: Find acceleration (retardation)

$$a = \frac{v - u}{t} = \frac{0 - 50}{0.01} = -5000 \text{ m s}^{-2}$$

Step 2: Find force using $F = ma$

$$F = ma = 0.5 \times (-5000) = -2500 \text{ N}$$

The negative sign indicates the force is directed opposite to the motion of the hammer (i.e., the nail exerts a retarding force on the hammer).

Answer: The force of the nail on the hammer is $2500$ N (directed opposite to the hammer's motion).

Given:
- Mass of motorcar, $m = 1200$ kg
- Initial velocity, $u = 90$ km/h $= 90 \times \frac{1000}{3600} = 25$ m s$^{-1}$
- Final velocity, $v = 18$ km/h $= 18 \times \frac{1000}{3600} = 5$ m s$^{-1}$
- Time, $t = 4$ s

Step 1: Calculate acceleration

$$a = \frac{v - u}{t} = \frac{5 - 25}{4} = \frac{-20}{4} = -5 \text{ m s}^{-2}$$

(Negative sign indicates retardation/deceleration)

Step 2: Calculate change in momentum

$$\Delta p = m(v - u) = 1200 \times (5 - 25) = 1200 \times (-20) = -24000 \text{ kg m s}^{-1}$$

Magnitude of change in momentum $= 24000$ kg m s$^{-1}$

Step 3: Calculate force required

$$F = ma = 1200 \times (-5) = -6000 \text{ N}$$

Magnitude of force $= 6000$ N

Answer:
- Acceleration $= -5$ m s$^{-2}$ (retardation of $5$ m s$^{-2}$)
- Change in momentum $= -24000$ kg m s$^{-1}$ (magnitude $= 24000$ kg m s$^{-1}$)
- Force required $= 6000$ N (opposing the motion)