Force and Laws of Motion — NCERT Solutions

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)